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== 11.4#h(0.6em)Hyperbolas

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Solve: #math.equation(block: false, alt: "x squared equals 12 .")[$x^(2) = 12 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus 2 the square root of 3")[$x = ± 2 sqrt(3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Expand: #math.equation(block: false, alt: "open parenthesis x minus 4 close parenthesis squared .")[$attach(( x − 4 ), t: 2) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x squared minus 8 x plus 16")[$x^(2) − 8 x + 16$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "y equals minus the fraction 2 over 3 x .")[$y = − frac(2, 3) x .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#figure(figph[A coordinate plane with a line graph passing through the origin (0,0), showing a negative slope, extending infinitely in both directions.], alt: "A coordinate plane with a line graph passing through the origin (0,0), showing a negative slope, extending infinitely in both directions.", caption: none)
]
]


=== Graph a Hyperbola with Center at #strong[(0, 0)]

The last conic section we will look at is called a #strong[hyperbola]. We will see that the equation of a hyperbola looks the same as the equation of an ellipse, except it is a difference rather than a sum. While the equations of an ellipse and a hyperbola are very similar, their graphs are very different.

We define a #strong[hyperbola] as all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a #strong[focus] of the hyperbola.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Hyperbola]

A #strong[hyperbola] is all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a #strong[focus] of the hyperbola.

#figure(figph[The figure shows a double napped right circular cone sliced by a plane that is parallel to the vertical axis of the cone forming a hyperbola. The figure is labeled ‘hyperbola’.], alt: "The figure shows a double napped right circular cone sliced by a plane that is parallel to the vertical axis of the cone forming a hyperbola. The figure is labeled ‘hyperbola’.", caption: none)
]

The line through the foci, is called the #strong[transverse axis]. The two points where the transverse axis intersects the hyperbola are each a #strong[vertex] of the hyperbola. The midpoint of the segment joining the foci is called the #strong[center] of the hyperbola. The line perpendicular to the transverse axis that passes through the center is called the #strong[conjugate axis]. Each piece of the graph is called a #strong[branch] of the hyperbola.

#figure(figph[The figure shows two graphs of a hyperbola. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci are shown with points that lie on the transverse axis, which is the x-axis. The branches pass through the vertices and open left and right. The y-axis is the conjugate axis. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci lie are shown with points that lie on the transverse axis, which is the y-axis. The branches pass through the vertices and open up and down. The x-axis is the conjugate axis.], alt: "The figure shows two graphs of a hyperbola. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci are shown with points that lie on the transverse axis, which is the x-axis. The branches pass through the vertices and open left and right. The y-axis is the conjugate axis. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci lie are shown with points that lie on the transverse axis, which is the y-axis. The branches pass through the vertices and open up and down. The x-axis is the conjugate axis.", caption: none)

Again our goal is to connect the geometry of a conic with algebra. Placing the hyperbola on a rectangular coordinate system gives us that opportunity. In the figure, we placed the hyperbola so the foci #math.equation(block: false, alt: "open parenthesis open parenthesis − c , 0 close parenthesis , open parenthesis c , 0 close parenthesis close parenthesis")[$( ( "−" c , 0 ) , ( c , 0 ) )$] are on the #emph[x]-axis and the center is the origin.

#figure(figph[The figure shows the graph of a hyperbola. The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The foci (negative c, 0) and (c, 0) are marked with a point and lie on the x-axis. The vertices are marked with a point and lie on the x-axis. The branches pass through the vertices and open left and right. The distance from (negative c, 0) to a point on the branch (x, y) is marked d sub 1. The distance from (x, y) on the branch to (c, 0) is marked d sub 2.], alt: "The figure shows the graph of a hyperbola. The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The foci (negative c, 0) and (c, 0) are marked with a point and lie on the x-axis. The vertices are marked with a point and lie on the x-axis. The branches pass through the vertices and open left and right. The distance from (negative c, 0) to a point on the branch (x, y) is marked d sub 1. The distance from (x, y) on the branch to (c, 0) is marked d sub 2.", caption: none)

The definition states the difference of the distance from the foci to a point #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] is constant. So #math.equation(block: false, alt: "| d sub 1 minus d sub 2 |")[$| d_(1) − d_(2) |$] is a constant that we will call #math.equation(block: false, alt: "2 a")[$2 a$] so #math.equation(block: false, alt: "| d sub 1 minus d sub 2 | equals 2 a .")[$| d_(1) − d_(2) | = 2 a .$] We will use the distance formula to lead us to an algebraic formula for an ellipse.

#math.equation(block: true, alt: "| d sub 1 minus d sub 2 | equals 2 a; Use the distance formula to find d sub 1 , d sub 2, | the square root of open parenthesis x minus open parenthesis minus c close parenthesis close parenthesis squared plus open parenthesis y minus 0 close parenthesis squared minus the square root of open parenthesis x minus c close parenthesis squared plus open parenthesis y minus 0 close parenthesis squared | equals 2 a; Eliminate the radicals.; To simplify the equation of the ellipse, we; let c squared minus a squared equals b squared ., the fraction x squared over a squared plus the fraction y squared over c squared minus a squared equals 1 ; So, the equation of a hyperbola centered at; the origin in standard form is:, the fraction x squared over a squared minus the fraction y squared over b squared")[$&  &  & #h(4em) | d_(1) #h(3.5em) − #h(3.5em) d_(2) | #h(3.2em) = 2 a \ "Use the distance formula to find" #h(0.2em) d_(1) , d_(2) &  &  & #h(2em) | sqrt(attach(( x − ( − c ) ), t: 2) + attach(( y − 0 ), t: 2)) − sqrt(attach(( x − c ), t: 2) + attach(( y − 0 ), t: 2)) #h(0.2em) | = 2 a \ "Eliminate the radicals." &  &  &  \ "To simplify the equation of the ellipse, we" \ "let" #h(0.2em) c^(2) − a^(2) = b^(2) . &  &  & #h(2em) frac(x^(2), a^(2)) + frac(y^(2), c^(2) − a^(2)) = 1 #h(0.53em) \ "So, the equation of a hyperbola centered at" \ "the origin in standard form is:" &  &  & #h(2em) frac(x^(2), a^(2)) − frac(y^(2), b^(2)) = 1 #h(0.53em)$]

To graph the hyperbola, it will be helpful to know about the intercepts. We will find the #emph[x]-intercepts and #emph[y]-intercepts using the formula.

#math.equation(block: true, alt: "x -intercepts, y -intercepts; the fraction x squared over a squared minus the fraction y squared over b squared, equals, 1, the fraction x squared over a squared minus the fraction y squared over b squared, equals, 1; Let y equals 0 ., the fraction x squared over a squared minus the fraction 0 squared over b squared, equals, 1, Let x equals 0 ., the fraction 0 squared over a squared minus the fraction y squared over b squared, equals, 1; the fraction x squared over a squared, equals, 1, minus the fraction y squared over b squared, equals, 1; x squared, equals, a squared, y squared, equals, − b squared; x, equals, ± a, y, equals, ± the square root of − b squared; The x -intercepts are open parenthesis a , 0 close parenthesis and open parenthesis − a , 0 close parenthesis ., There are no y -")[$&  &  &  &  & #h(5em) "x" "-intercepts" &  &  &  &  &  &  &  &  &  & #h(9em) "y" "-intercepts" \  &  &  &  &  & frac(x^(2), a^(2)) − frac(y^(2), b^(2)) & = & 1 &  &  &  &  &  &  &  &  &  & frac(x^(2), a^(2)) − frac(y^(2), b^(2)) & = & 1 \ "Let" #h(0.2em) y = 0 . &  &  &  &  & frac(x^(2), a^(2)) − frac(0^(2), b^(2)) & = & 1 &  &  &  &  & #h(3em) "Let" #h(0.2em) x = 0 . &  &  &  &  & frac(0^(2), a^(2)) − frac(y^(2), b^(2)) & = & 1 \  &  &  &  &  & frac(x^(2), a^(2)) & = & 1 &  &  &  &  &  &  &  &  &  & − frac(y^(2), b^(2)) & = & 1 \  &  &  &  &  & x^(2) & = & a^(2) &  &  &  &  &  &  &  &  &  & y^(2) & = & "−" b^(2) \  &  &  &  &  & x & = & "±" a &  &  &  &  &  &  &  &  &  & y & = & "±" sqrt("−" b^(2)) \ #h(2em) "The" #h(0.2em) x "-intercepts are" #h(0.2em) ( a , 0 ) #h(0.2em) "and" #h(0.2em) ( "−" a , 0 ) . &  &  &  &  & #h(5em) "There are no" #h(0.2em) y "-intercepts."$]

The #emph[a], #emph[b] values in the equation also help us find the asymptotes of the hyperbola. The asymptotes are intersecting straight lines that the branches of the graph approach but never intersect as the #emph[x], #emph[y] values get larger and larger.

To find the asymptotes, we sketch a rectangle whose sides intersect the #emph[x]-axis at the vertices #math.equation(block: false, alt: "open parenthesis − a , 0 close parenthesis ,")[$( "−" a , 0 ) ,$] #math.equation(block: false, alt: "open parenthesis a , 0 close parenthesis")[$( a , 0 )$] and intersect the #emph[y]-axis at #math.equation(block: false, alt: "open parenthesis 0 , − b close parenthesis ,")[$( 0 , "−" b ) ,$] #math.equation(block: false, alt: "open parenthesis 0 , b close parenthesis .")[$( 0 , b ) .$] The lines containing the diagonals of this rectangle are the asymptotes of the hyperbola. The rectangle and asymptotes are not part of the hyperbola, but they help us graph the hyperbola.

#figure(figph[The figure shows the graph of a hyperbola. The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (negative a, 0) and (a, 0) and are marked with a point and lie on the x-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the x-axis at the vertices (negative a, 0) and (a, 0) and intersect the y-axis at (0, b) and (0, negative b). The asymptotes are given by y is equal to b divided by a times x and y is equal to negative b divided by a times x and are drawn as the diagonals of the central rectangle. The branches of the hyperbola pass through the vertices, open left and right, and approach the asymptotes.], alt: "The figure shows the graph of a hyperbola. The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (negative a, 0) and (a, 0) and are marked with a point and lie on the x-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the x-axis at the vertices (negative a, 0) and (a, 0) and intersect the y-axis at (0, b) and (0, negative b). The asymptotes are given by y is equal to b divided by a times x and y is equal to negative b divided by a times x and are drawn as the diagonals of the central rectangle. The branches of the hyperbola pass through the vertices, open left and right, and approach the asymptotes.", caption: none)

The asymptotes pass through the origin and we can evaluate their slope using the rectangle we sketched. They have equations #math.equation(block: false, alt: "y equals the fraction b over a x")[$y = frac(b, a) x$] and #math.equation(block: false, alt: "y equals minus the fraction b over a x .")[$y = − frac(b, a) x .$]

There are two equations for hyperbolas, depending whether the transverse axis is vertical or horizontal. We can tell whether the transverse axis is horizontal by looking at the equation. When the equation is in standard form, if the #emph[x]#super[2]-term is positive, the transverse axis is horizontal. When the equation is in standard form, if the #emph[y]#super[2]-term is positive, the transverse axis is vertical.

The second equations could be derived similarly to what we have done. We will summarize the results here.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Standard Form of the Equation a Hyperbola with Center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]]

The standard form of the equation of a hyperbola with center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis ,")[$( 0 , 0 ) ,$] is

#math.equation(block: true, alt: "the fraction x squared over a squared minus the fraction y squared over b squared equals 1 or the fraction y squared over a squared minus the fraction x squared over b squared equals 1")[$frac(x^(2), a^(2)) − frac(y^(2), b^(2)) = 1 #h(2em) "or" #h(2em) frac(y^(2), a^(2)) − frac(x^(2), b^(2)) = 1$]#figure(figph[The figure shows the graph of two hyperbolas. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (negative a, 0) and (a, 0) and are marked with a point and lie on the x-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the x-axis at the vertices (negative a, 0) and (a, 0) and intersect the y-axis at (0, b) and (0, negative b). The asymptotes are given by y is equal to b divided by a times x and y is equal to negative b divided by a times x and are drawn as the diagonals of the central rectangle. The branches of the hyperbola pass through the vertices, open left and right, and approach the asymptotes. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (0, a) and (0, negative a) and are marked with a point and lie on the y-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the y-axis at the vertices (0, a) and (0, negative a) and intersect the y-axis at (negative b, 0) and (b, 0). The branches of the hyperbola pass through the vertices, open up and down, and approach the asymptotes.], alt: "The figure shows the graph of two hyperbolas. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (negative a, 0) and (a, 0) and are marked with a point and lie on the x-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the x-axis at the vertices (negative a, 0) and (a, 0) and intersect the y-axis at (0, b) and (0, negative b). The asymptotes are given by y is equal to b divided by a times x and y is equal to negative b divided by a times x and are drawn as the diagonals of the central rectangle. The branches of the hyperbola pass through the vertices, open left and right, and approach the asymptotes. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (0, a) and (0, negative a) and are marked with a point and lie on the y-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the y-axis at the vertices (0, a) and (0, negative a) and intersect the y-axis at (negative b, 0) and (b, 0). The branches of the hyperbola pass through the vertices, open up and down, and approach the asymptotes.", caption: none)

Notice that, unlike the equation of an ellipse, the denominator of #math.equation(block: false, alt: "x squared")[$x^(2)$] is not always #math.equation(block: false, alt: "a squared")[$a^(2)$] and the denominator of #math.equation(block: false, alt: "y squared")[$y^(2)$] is not always #math.equation(block: false, alt: "b squared .")[$b^(2) .$]

Notice that when the #math.equation(block: false, alt: "x squared")[$x^(2)$]-term is positive, the transverse axis is on the #emph[x]-axis. When the #math.equation(block: false, alt: "y squared")[$y^(2)$]-term is positive, the transverse axis is on the #emph[y]-axis.
]

#figure(table(
  columns: 3, align: left, inset: 6pt,
  [Standard Forms of the Equation a Hyperbola with Center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]], [], [],
  [], [#math.equation(block: false, alt: "the fraction x squared over a squared minus the fraction y squared over b squared equals 1")[$frac(x^(2), a^(2)) − frac(y^(2), b^(2)) = 1$]], [#math.equation(block: false, alt: "the fraction y squared over a squared minus the fraction x squared over b squared equals 1")[$frac(y^(2), a^(2)) − frac(x^(2), b^(2)) = 1$]],
  [#strong[Orientation]], [Transverse axis on the #emph[x]-axis. #linebreak() Opens left and right], [Transverse axis on the #emph[y]-axis. #linebreak() Opens up and down],
  [#strong[Vertices]], [#math.equation(block: false, alt: "open parenthesis − a , 0 close parenthesis ,")[$( "−" a , 0 ) ,$] #math.equation(block: false, alt: "open parenthesis a , 0 close parenthesis")[$( a , 0 )$]], [#math.equation(block: false, alt: "open parenthesis 0 , − a close parenthesis ,")[$( 0 , "−" a ) ,$] #math.equation(block: false, alt: "open parenthesis 0 , a close parenthesis")[$( 0 , a )$]],
  [#strong[#emph[x]-intercepts]], [#math.equation(block: false, alt: "open parenthesis − a , 0 close parenthesis ,")[$( "−" a , 0 ) ,$] #math.equation(block: false, alt: "open parenthesis a , 0 close parenthesis")[$( a , 0 )$]], [none],
  [#strong[#emph[y]-intercepts]], [none], [#math.equation(block: false, alt: "open parenthesis 0 , − a close parenthesis ,")[$( 0 , "−" a ) ,$] #math.equation(block: false, alt: "open parenthesis 0 , a close parenthesis")[$( 0 , a )$]],
  [#strong[Rectangle]], [Use #math.equation(block: false, alt: "open parenthesis ± a , 0 close parenthesis")[$( "±" a , 0 )$] #math.equation(block: false, alt: "open parenthesis 0 , ± b close parenthesis")[$( 0 , "±" b )$]], [Use #math.equation(block: false, alt: "open parenthesis 0 , ± a close parenthesis")[$( 0 , "±" a )$] #math.equation(block: false, alt: "open parenthesis ± b , 0 close parenthesis")[$( "±" b , 0 )$]],
  [#strong[asymptotes]], [#math.equation(block: false, alt: "y equals the fraction b over a x ,")[$y = frac(b, a) x ,$] #math.equation(block: false, alt: "y equals minus the fraction b over a x")[$y = − frac(b, a) x$]], [#math.equation(block: false, alt: "y equals the fraction a over b x ,")[$y = frac(a, b) x ,$] #math.equation(block: false, alt: "y equals minus the fraction a over b x")[$y = − frac(a, b) x$]],
))

We will use these properties to graph hyperbolas.

#examplebox("Example 1")[How to Graph a Hyperbola with Center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]][
Graph #math.equation(block: false, alt: "the fraction x squared over 25 minus the fraction y squared over 4 equals 1 .")[$frac(x^(2), 25) − frac(y^(2), 4) = 1 .$]

#solutionbox[
#figure(figph[Step 1 is to write the equation in standard form. The the quantity x squared divided by 25 end quantity minus the quantity y squared divided by 4 end quantity is equal to 1 is already in standard form.], alt: "Step 1 is to write the equation in standard form. The the quantity x squared divided by 25 end quantity minus the quantity y squared divided by 4 end quantity is equal to 1 is already in standard form.", caption: none)

#figure(figph[Step 2 is to determine whether the transverse axis is horizontal or vertical. Since the x squared term is positive, the transverse axis is horizontal.], alt: "Step 2 is to determine whether the transverse axis is horizontal or vertical. Since the x squared term is positive, the transverse axis is horizontal.", caption: none)

#figure(figph[Step 3 is to find the vertices. Since a squared is equal to 25, then a is equal to plus or minus 5. The vertices lie on the x-axis and are (negative 5, 0) and (5, 0).], alt: "Step 3 is to find the vertices. Since a squared is equal to 25, then a is equal to plus or minus 5. The vertices lie on the x-axis and are (negative 5, 0) and (5, 0).", caption: none)

#figure(figph[Step 4 is to sketch the rectangle centered at the origin, intersecting one axis at plus or minus a and the other at plus or minus b. Since a is equal to plus or minus 5, the rectangle will intersect the x-axis at the vertices. Since b is equal to plus or minus 2, the rectangle will intersect the y-axis at (0, negative 2) and (0, 2). The rectangle is shown on a coordinate plane with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled.], alt: "Step 4 is to sketch the rectangle centered at the origin, intersecting one axis at plus or minus a and the other at plus or minus b. Since a is equal to plus or minus 5, the rectangle will intersect the x-axis at the vertices. Since b is equal to plus or minus 2, the rectangle will intersect the y-axis at (0, negative 2) and (0, 2). The rectangle is shown on a coordinate plane with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled.", caption: none)

#figure(figph[Step 5 is to sketch the asymptotes, the lines through the diagonals of the rectangle. The asymptotes have the equations y is equal to five-halves times x and y is equal to negative five-halves x. The coordinate plane shows the rectangle with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled and the lines that represent the asymptotes.], alt: "Step 5 is to sketch the asymptotes, the lines through the diagonals of the rectangle. The asymptotes have the equations y is equal to five-halves times x and y is equal to negative five-halves x. The coordinate plane shows the rectangle with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled and the lines that represent the asymptotes.", caption: none)

#figure(figph[Step 6 is to draw the two branches of the hyperbola. Start at each vertex and use the asymptotes as a guide. The coordinate plane shows the rectangle with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled, the lines that represent the asymptotes, y is equal to plus or minus five-halves times x, and the branches that pass through (plus or minus 5, 0) and open left and right.], alt: "Step 6 is to draw the two branches of the hyperbola. Start at each vertex and use the asymptotes as a guide. The coordinate plane shows the rectangle with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled, the lines that represent the asymptotes, y is equal to plus or minus five-halves times x, and the branches that pass through (plus or minus 5, 0) and open left and right.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "the fraction x squared over 16 minus the fraction y squared over 4 equals 1 .")[$frac(x^(2), 16) − frac(y^(2), 4) = 1 .$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus one-half times x, and branches that pass through the vertices (plus or minus 4, 0) and open left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus one-half times x, and branches that pass through the vertices (plus or minus 4, 0) and open left and right.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "the fraction x squared over 9 minus the fraction y squared over 16 equals 1 .")[$frac(x^(2), 9) − frac(y^(2), 16) = 1 .$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus four-thirds times x, and branches that pass through the vertices (plus or minus 3, 0) and open left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus four-thirds times x, and branches that pass through the vertices (plus or minus 3, 0) and open left and right.", caption: none)
]
]

We summarize the steps for reference.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Graph a hyperbola centered at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$]]

+ Write the equation in standard form.
+ Determine whether the transverse axis is horizontal or vertical.
+ Find the vertices.
+ Sketch the rectangle centered at the origin intersecting one axis at #math.equation(block: false, alt: "± a")[$"±" a$] and the other at #math.equation(block: false, alt: "± b .")[$"±" b .$]
+ Sketch the asymptotes—the lines through the diagonals of the rectangle.
+ Draw the two branches of the hyperbola.
]

Sometimes the equation for a hyperbola needs to be first placed in standard form before we graph it.

#examplebox("Example 2")[][
Graph #math.equation(block: false, alt: "4 y squared minus 16 x squared equals 64 .")[$4 y^(2) − 16 x^(2) = 64 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "4 y squared minus 16 x squared equals 64")[$4 y^(2) − 16 x^(2) = 64$]]),
  [To write the equation in standard form, divide #linebreak() each term by 64 to make the equation equal to 1.], [#math.equation(block: false, alt: "the fraction 4 y squared over 64 minus the fraction 16 x squared over 64 equals the fraction 64 over 64")[$frac(4 y^(2), 64) − frac(16 x^(2), 64) = frac(64, 64)$]],
  [Simplify.], [#math.equation(block: false, alt: "the fraction y squared over 16 minus the fraction x squared over 4 equals 1")[$#h(1em) frac(y^(2), 16) − frac(x^(2), 4) = 1$]],
  [Since the #emph[y]#super[2]-term is positive, the transverse axis is vertical. #linebreak() Since #math.equation(block: false, alt: "a squared equals 16")[$a^(2) = 16$] then #math.equation(block: false, alt: "a equals ± 4 .")[$a = "±" 4 .$]], [],
  [The vertices are on the #emph[y]-axis, #math.equation(block: false, alt: "open parenthesis 0 , − a close parenthesis ,")[$( 0 , "−" a ) ,$] #math.equation(block: false, alt: "open parenthesis 0 , a close parenthesis .")[$( 0 , a ) .$] #linebreak() Since #math.equation(block: false, alt: "b squared equals 4")[$b^(2) = 4$] then #math.equation(block: false, alt: "b equals ± 2 .")[$b = "±" 2 .$]], [#math.equation(block: false, alt: "open parenthesis 0 , −4 close parenthesis ,")[$( 0 , −4 ) ,$] #math.equation(block: false, alt: "open parenthesis 0 , 4 close parenthesis")[$( 0 , 4 )$]],
  [Sketch the rectangle intersecting the #emph[x]-axis at #math.equation(block: false, alt: "open parenthesis −2 , 0 close parenthesis ,")[$( −2 , 0 ) ,$] #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$] and the #emph[y]-axis at the vertices. #linebreak() Sketch the asymptotes through the diagonals of the rectangle. #linebreak() Draw the two branches of the hyperbola.], [#figure(figph[Hyperbola centered at the origin with vertices at (0,4) and (0,-4), showing asymptotes and the reference rectangle.], alt: "Hyperbola centered at the origin with vertices at (0,4) and (0,-4), showing asymptotes and the reference rectangle.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "4 y squared minus 25 x squared equals 100 .")[$4 y^(2) − 25 x^(2) = 100 .$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus five-halves times x, and branches that pass through the vertices (0, plus or minus 5) and open up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus five-halves times x, and branches that pass through the vertices (0, plus or minus 5) and open up and down.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "25 y squared minus 9 x squared equals 225 .")[$25 y^(2) − 9 x^(2) = 225 .$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus three-fifths times x, and branches that pass through the vertices (0, plus or minus 3) and open up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus three-fifths times x, and branches that pass through the vertices (0, plus or minus 3) and open up and down.", caption: none)
]
]


=== Graph a Hyperbola with Center at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]

Hyperbolas are not always centered at the origin. When a hyperbola is centered at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$] the equations changes a bit as reflected in the table.

#figure(table(
  columns: 3, align: left, inset: 6pt,
  [Standard Forms of the Equation a Hyperbola with Center #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]], [], [],
  [], [#math.equation(block: false, alt: "the fraction open parenthesis x minus h close parenthesis squared over a squared minus the fraction open parenthesis y minus k close parenthesis squared over b squared equals 1")[$frac(attach(( x − h ), t: 2), a^(2)) − frac(attach(( y − k ), t: 2), b^(2)) = 1$]], [#math.equation(block: false, alt: "the fraction open parenthesis y minus k close parenthesis squared over a squared minus the fraction open parenthesis x minus h close parenthesis squared over b squared equals 1")[$frac(attach(( y − k ), t: 2), a^(2)) − frac(attach(( x − h ), t: 2), b^(2)) = 1$]],
  [#strong[Orientation]], [Transverse axis is horizontal. #linebreak() Opens left and right], [Transverse axis is vertical. #linebreak() Opens up and down],
  [#strong[Center]], [#math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]], [#math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]],
  [#strong[Vertices]], [#emph[a] units to the left and right of the center], [#emph[a] units above and below the center],
  [#strong[Rectangle]], [Use #emph[a] units left/right of center #linebreak() #emph[b] units above/ below the center], [Use #emph[a] units above/below the center #linebreak() #emph[b] units left/right of center],
))

#examplebox("Example 3")[How to Graph a Hyperbola with Center #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]][
Graph #math.equation(block: false, alt: "the fraction open parenthesis x minus 1 close parenthesis squared over 9 minus the fraction open parenthesis y minus 2 close parenthesis squared over 16 equals 1")[$frac(attach(( x − 1 ), t: 2), 9) − frac(attach(( y − 2 ), t: 2), 16) = 1$]

#solutionbox[
#figure(figph[Step 1 is to write the equation in standard form. Notice that the equation the quantity x minus 1 squared all divided by 9 end quantity minus the quantity y minus 2 squared all divided by 16 end quantity is equal to 1 is already in standard form.], alt: "Step 1 is to write the equation in standard form. Notice that the equation the quantity x minus 1 squared all divided by 9 end quantity minus the quantity y minus 2 squared all divided by 16 end quantity is equal to 1 is already in standard form.", caption: none)

#figure(figph[Step 2 is to deteremine whether the transverse axis is horizonal or vertical. Since the x squared term is positive, the hyperbola opens left and right. The transverse axis is horizontal. The hyperbola opens left and right.], alt: "Step 2 is to deteremine whether the transverse axis is horizonal or vertical. Since the x squared term is positive, the hyperbola opens left and right. The transverse axis is horizontal. The hyperbola opens left and right.", caption: none)

#figure(figph[Step 3 is to find the center and a and b. h is equal to 1 and k is equal 2. a squared is equal to 9 and b squared is equal to 16. You can see tha x minus h is x minus 1, and that y minus k is y minus 2. So, the center is (1, 2) and a is equal to 3 and b is equal to 4.], alt: "Step 3 is to find the center and a and b. h is equal to 1 and k is equal 2. a squared is equal to 9 and b squared is equal to 16. You can see tha x minus h is x minus 1, and that y minus k is y minus 2. So, the center is (1, 2) and a is equal to 3 and b is equal to 4.", caption: none)

#figure(figph[Step 4 is to sketch the rectangle centered at (h, k) using a and b. Mark the center (1, 2) on a coordinate plane. Sketch the rectangle that goes through the points 3 units to the left and right of the center and 4 units above and below the center.], alt: "Step 4 is to sketch the rectangle centered at (h, k) using a and b. Mark the center (1, 2) on a coordinate plane. Sketch the rectangle that goes through the points 3 units to the left and right of the center and 4 units above and below the center.", caption: none)

#figure(figph[Step 5 is to sketch the asymptotes on the coordinate plane. They are the lines through the diagonals of the retcangle. Mark the vertices which lie on the rectangle 3 units to the left and right of the center. The vertices are (negative 2, 2) and (4, 2).], alt: "Step 5 is to sketch the asymptotes on the coordinate plane. They are the lines through the diagonals of the retcangle. Mark the vertices which lie on the rectangle 3 units to the left and right of the center. The vertices are (negative 2, 2) and (4, 2).", caption: none)

#figure(figph[Step 6 is to draw the branches of the hyperbola. Start the vertices, (negative 2, 2) and (4, 2) and use the asymptotes as a guide. The branches should open left and right.], alt: "Step 6 is to draw the branches of the hyperbola. Start the vertices, (negative 2, 2) and (4, 2) and use the asymptotes as a guide. The branches should open left and right.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "the fraction open parenthesis x minus 3 close parenthesis squared over 25 minus the fraction open parenthesis y minus 1 close parenthesis squared over 9 equals 1 .")[$frac(attach(( x − 3 ), t: 2), 25) − frac(attach(( y − 1 ), t: 2), 9) = 1 .$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with an asymptote that passes through (negative 2, negative 2) and (8, 4) and an asymptote that passes through (negative 2, 4) and (8, negative 2), and branches that pass through the vertices (negative 2, 2) and (8, 2) and opens left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with an asymptote that passes through (negative 2, negative 2) and (8, 4) and an asymptote that passes through (negative 2, 4) and (8, negative 2), and branches that pass through the vertices (negative 2, 2) and (8, 2) and opens left and right.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "the fraction open parenthesis x minus 2 close parenthesis squared over 4 minus the fraction open parenthesis y minus 2 close parenthesis squared over 9 equals 1 .")[$frac(attach(( x − 2 ), t: 2), 4) − frac(attach(( y − 2 ), t: 2), 9) = 1 .$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (2, 2), an asymptote that passes through (0, negative 1) and (4, 5) and an asymptote that passes through (0, 5) and (4, negative 1), and branches that pass through the vertices (0, 2) and (4, 2) and opens left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (2, 2), an asymptote that passes through (0, negative 1) and (4, 5) and an asymptote that passes through (0, 5) and (4, negative 1), and branches that pass through the vertices (0, 2) and (4, 2) and opens left and right.", caption: none)
]
]

We summarize the steps for easy reference.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Graph a hyperbola centered at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis .")[$( h , k ) .$]]

+ Write the equation in standard form.
+ Determine whether the transverse axis is horizontal or vertical.
+ Find the center and #emph[a, b].
+ Sketch the rectangle centered at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$] using #emph[a, b].
+ Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices.
+ Draw the two branches of the hyperbola.
]

Be careful as you identify the center. The standard equation has #math.equation(block: false, alt: "x minus h")[$x − h$] and #math.equation(block: false, alt: "y minus k")[$y − k$] with the center as #math.equation(block: false, alt: "open parenthesis h , k close parenthesis .")[$( h , k ) .$]

#examplebox("Example 4")[][
Graph #math.equation(block: false, alt: "the fraction open parenthesis y plus 2 close parenthesis squared over 9 minus the fraction open parenthesis x plus 1 close parenthesis squared over 4 equals 1 .")[$frac(attach(( y + 2 ), t: 2), 9) − frac(attach(( x + 1 ), t: 2), 4) = 1 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation for a hyperbola is displayed, reading as (y+2)^2 / 9 - (x+1)^2 / 4 = 1. The equation shows two squared binomials over numerical denominators, set equal to one.], alt: "A mathematical equation for a hyperbola is displayed, reading as (y+2)^2 / 9 - (x+1)^2 / 4 = 1. The equation shows two squared binomials over numerical denominators, set equal to one.", caption: none)]),
  [Since the #math.equation(block: false, alt: "y squared -")[$y^(2) "-"$]term is positive, the hyperbola #linebreak() opens up and down.], [#figure(figph[A mathematical equation for a hyperbola is displayed, with \\( (y - (-2))^2 \\) over 9 minus \\( (x - (-1))^2 \\) over 4, equaling 1. The variables 'k' and 'h' are shown in red above -(-2) and -(-1), respectively.], alt: "A mathematical equation for a hyperbola is displayed, with \\( (y - (-2))^2 \\) over 9 minus \\( (x - (-1))^2 \\) over 4, equaling 1. The variables 'k' and 'h' are shown in red above -(-2) and -(-1), respectively.", caption: none)],
  [Find the center, #math.equation(block: false, alt: "open parenthesis h , k close parenthesis .")[$( h , k ) .$]], [Center: #math.equation(block: false, alt: "open parenthesis −1 , −2 close parenthesis")[$( −1 , −2 )$]],
  [Find #emph[a, b].], [#math.equation(block: false, alt: "a equals 3")[$a = 3$] #math.equation(block: false, alt: "b equals 2")[$b = 2$]],
  [Sketch the rectangle that goes through the #linebreak() points 3 units above and below the center and #linebreak() 2 units to the left/right of the center. #linebreak() Sketch the asymptotes—the lines through the #linebreak() diagonals of the rectangle. #linebreak() Mark the vertices. #linebreak() Graph the branches.], [#figure(figph[A graph displays a hyperbola centered at (-1,-2) on a coordinate plane. The hyperbola opens vertically, with its two branches extending upwards and downwards. Dashed blue lines represent the asymptotes, and a dashed red rectangle indicates the central box.], alt: "A graph displays a hyperbola centered at (-1,-2) on a coordinate plane. The hyperbola opens vertically, with its two branches extending upwards and downwards. Dashed blue lines represent the asymptotes, and a dashed red rectangle indicates the central box.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "the fraction open parenthesis y plus 3 close parenthesis squared over 16 minus the fraction open parenthesis x plus 2 close parenthesis squared over 9 equals 1 .")[$frac(attach(( y + 3 ), t: 2), 16) − frac(attach(( x + 2 ), t: 2), 9) = 1 .$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with a center at (negative 2, negative 3), an asymptote that passes through (negative 5, negative 7) and (1, 1) and an asymptote that passes through (negative 5, 1) and (1, 7), and branches that pass through the vertices (negative 2, 1) and (negative 2, negative 7) and opens up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with a center at (negative 2, negative 3), an asymptote that passes through (negative 5, negative 7) and (1, 1) and an asymptote that passes through (negative 5, 1) and (1, 7), and branches that pass through the vertices (negative 2, 1) and (negative 2, negative 7) and opens up and down.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "the fraction open parenthesis y plus 2 close parenthesis squared over 9 minus the fraction open parenthesis x plus 2 close parenthesis squared over 9 equals 1 .")[$frac(attach(( y + 2 ), t: 2), 9) − frac(attach(( x + 2 ), t: 2), 9) = 1 .$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with a center at (negative 2, negative 2), an asymptote that passes through (negative 5, negative 5) and (1, 1) and an asymptote that passes through (negative 5, 1) and (1, negative 5), and branches that pass through the vertices (negative 2, 1) and (negative 2, negative 5) and opens up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with a center at (negative 2, negative 2), an asymptote that passes through (negative 5, negative 5) and (1, 1) and an asymptote that passes through (negative 5, 1) and (1, negative 5), and branches that pass through the vertices (negative 2, 1) and (negative 2, negative 5) and opens up and down.", caption: none)
]
]

Again, sometimes we have to put the equation in standard form as our first step.

#examplebox("Example 5")[][
Write the equation in standard form and graph #math.equation(block: false, alt: "4 x squared minus 9 y squared minus 24 x minus 36 y minus 36 equals 0 .")[$4 x^(2) − 9 y^(2) − 24 x − 36 y − 36 = 0 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed on a white background: 4x^2 - 9y^2 - 24x - 36y - 36 = 0.], alt: "A mathematical equation is displayed on a white background: 4x^2 - 9y^2 - 24x - 36y - 36 = 0.", caption: none)]),
  [To get to standard form, complete the squares.], [#figure(figph[The image shows the mathematical equation 4(x^2 - 6x) - 9(y^2 + 4y) = 36, which is an equation of a hyperbola.], alt: "The image shows the mathematical equation 4(x^2 - 6x) - 9(y^2 + 4y) = 36, which is an equation of a hyperbola.", caption: none)],
  [], [#figure(figph[An algebraic equation: 4(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 36 + 36 - 36, illustrating a step in completing the square for a conic section, with specific constants highlighted.], alt: "An algebraic equation: 4(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 36 + 36 - 36, illustrating a step in completing the square for a conic section, with specific constants highlighted.", caption: none)],
  [], [#figure(figph[The image displays the mathematical equation 4(x-3)^2 - 9(y+2)^2 = 36, which is the standard form of a hyperbola.], alt: "The image displays the mathematical equation 4(x-3)^2 - 9(y+2)^2 = 36, which is the standard form of a hyperbola.", caption: none)],
  [Divide each term by 36 to get the constant to be 1.], [#figure(figph[An equation for a hyperbola is displayed, showing the difference of two fractions: 4(x-3)^2/36 minus 9(y+2)^2/36, which equals 36/36.], alt: "An equation for a hyperbola is displayed, showing the difference of two fractions: 4(x-3)^2/36 minus 9(y+2)^2/36, which equals 36/36.", caption: none)],
  [], [#figure(figph[A mathematical equation for a hyperbola is displayed, reading: (x-3)^2 / 9 - (y+2)^2 / 4 = 1.], alt: "A mathematical equation for a hyperbola is displayed, reading: (x-3)^2 / 9 - (y+2)^2 / 4 = 1.", caption: none)],
  [Since the #math.equation(block: false, alt: "x squared -")[$x^(2) "-"$]term is positive, the hyperbola #linebreak() opens left and right.], [],
  [Find the center, #math.equation(block: false, alt: "open parenthesis h , k close parenthesis .")[$( h , k ) .$]], [Center: #math.equation(block: false, alt: "open parenthesis 3 , −2 close parenthesis")[$( 3 , −2 )$]],
  [Find #emph[a, b].], [#math.equation(block: false, alt: "a equals 3; b equals 4")[$a = 3 \ b = 4$]],
  [Sketch the rectangle that goes through the #linebreak() points 3 units to the left/right of the center #linebreak() and 2 units above and below the center. #linebreak() Sketch the asymptotes—the lines through the #linebreak() diagonals of the rectangle. #linebreak() Mark the vertices. #linebreak() Graph the branches.], [#figure(figph[This image depicts a hyperbola on a coordinate grid, with its center clearly marked at (3, -2). The hyperbola's branches extend outward, guided by its dashed asymptotes.], alt: "This image depicts a hyperbola on a coordinate grid, with its center clearly marked at (3, -2). The hyperbola's branches extend outward, guided by its dashed asymptotes.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Write the equation in standard form and ⓑ graph #math.equation(block: false, alt: "9 x squared minus 16 y squared plus 18 x plus 64 y minus 199 equals 0 .")[$9 x^(2) − 16 y^(2) + 18 x + 64 y − 199 = 0 .$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "the fraction open parenthesis x plus 1 close parenthesis squared over 16 minus the fraction open parenthesis y minus 2 close parenthesis squared over 9 equals 1")[$frac(attach(( x + 1 ), t: 2), 16) − frac(attach(( y − 2 ), t: 2), 9) = 1$] #linebreak() 
ⓑ  #linebreak() #figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (negative 1, 2), an asymptote that passes through (negative 5, 5) and (3, negative 1) and an asymptote that passes through (3, 5) and (negative 5, negative 1), and branches that pass through the vertices (negative 5, 2) and (3, 2) and opens left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (negative 1, 2), an asymptote that passes through (negative 5, 5) and (3, negative 1) and an asymptote that passes through (3, 5) and (negative 5, negative 1), and branches that pass through the vertices (negative 5, 2) and (3, 2) and opens left and right.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Write the equation in standard form and ⓑ graph #math.equation(block: false, alt: "16 x squared minus 25 y squared plus 96 x minus 50 y minus 281 equals 0 .")[$16 x^(2) − 25 y^(2) + 96 x − 50 y − 281 = 0 .$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "the fraction open parenthesis x plus 3 close parenthesis squared over 25 minus the fraction open parenthesis y plus 1 close parenthesis squared over 16 equals 1")[$frac(attach(( x + 3 ), t: 2), 25) − frac(attach(( y + 1 ), t: 2), 16) = 1$] #linebreak() 
ⓑ  #linebreak() #figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (negative 3, negative 1), an asymptote that passes through (negative 8, negative 5) and (2, 3) and an asymptote that passes through (negative 8, 3) and (2, negative 5), and branches that pass through the vertices (negative 8, negative 1) and (2, negative 1) and opens left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (negative 3, negative 1), an asymptote that passes through (negative 8, negative 5) and (2, 3) and an asymptote that passes through (negative 8, 3) and (2, negative 5), and branches that pass through the vertices (negative 8, negative 1) and (2, negative 1) and opens left and right.", caption: none)
]
]


=== Identify Conic Sections by their Equations

Now that we have completed our study of the conic sections, we will take a look at the different equations and recognize some ways to identify a conic by its equation. When we are given an equation to graph, it is helpful to identify the conic so we know what next steps to take.

To identify a conic from its equation, it is easier if we put the variable terms on one side of the equation and the constants on the other.

#figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([Conic], [Characteristics of #math.equation(block: false, alt: "x squared -")[$x^(2) "-"$] and #math.equation(block: false, alt: "y squared -")[$y^(2) "-"$] terms], [Example]),
  [#strong[Parabola]], [Either #math.equation(block: false, alt: "x squared")[$x^(2)$] OR #math.equation(block: false, alt: "y squared .")[$y^(2) .$] Only one variable is squared.], [#math.equation(block: false, alt: "x equals 3 y squared minus 2 y plus 1")[$x = 3 y^(2) − 2 y + 1$]],
  [#strong[Circle]], [#math.equation(block: false, alt: "x squared -")[$x^(2) "-"$] and #math.equation(block: false, alt: "y squared -")[$y^(2) "-"$] terms have the same coefficients], [#math.equation(block: false, alt: "x squared plus y squared equals 49")[$x^(2) + y^(2) = 49$]],
  [#strong[Ellipse]], [#math.equation(block: false, alt: "x squared -")[$x^(2) "-"$] and #math.equation(block: false, alt: "y squared -")[$y^(2) "-"$] terms have the #strong[same] sign, different coefficients], [#math.equation(block: false, alt: "4 x squared plus 25 y squared equals 100")[$4 x^(2) + 25 y^(2) = 100$]],
  [#strong[Hyperbola]], [#math.equation(block: false, alt: "x squared -")[$x^(2) "-"$] and #math.equation(block: false, alt: "y squared -")[$y^(2) "-"$] terms have #strong[different] signs, different coefficients], [#math.equation(block: false, alt: "25 y squared minus 4 x squared equals 100")[$25 y^(2) − 4 x^(2) = 100$]],
))

#examplebox("Example 6")[][
Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

ⓐ #math.equation(block: false, alt: "9 x squared plus 4 y squared plus 56 y plus 160 equals 0")[$9 x^(2) + 4 y^(2) + 56 y + 160 = 0$] ⓑ #math.equation(block: false, alt: "9 x squared minus 16 y squared plus 18 x plus 64 y minus 199 equals 0")[$9 x^(2) − 16 y^(2) + 18 x + 64 y − 199 = 0$] ⓒ #math.equation(block: false, alt: "x squared plus y squared minus 6 x minus 8 y equals 0")[$x^(2) + y^(2) − 6 x − 8 y = 0$] ⓓ #math.equation(block: false, alt: "y equals −2 x squared minus 4 x minus 5")[$y = −2 x^(2) − 4 x − 5$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "9 x squared plus 4 y squared plus 56 y plus 160 equals 0")[$9 x^(2) + 4 y^(2) + 56 y + 160 = 0$]]),
  [The #math.equation(block: false, alt: "x squared")[$x^(2)$]- and #math.equation(block: false, alt: "y squared")[$y^(2)$]-terms have the same sign and different coefficients.], [Ellipse],
))

#linebreak() 
ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "9 x squared minus 16 y squared plus 18 x plus 64 y minus 199 equals 0")[$9 x^(2) − 16 y^(2) + 18 x + 64 y − 199 = 0$]]),
  [The #math.equation(block: false, alt: "x squared")[$x^(2)$]- and #math.equation(block: false, alt: "y squared")[$y^(2)$]-terms have different signs and different coefficients.], [Hyperbola],
))

#linebreak() 
ⓒ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "x squared plus y squared minus 6 x minus 8 y equals 0")[$x^(2) + y^(2) − 6 x − 8 y = 0$]]),
  [The #math.equation(block: false, alt: "x squared")[$x^(2)$]- and #math.equation(block: false, alt: "y squared")[$y^(2)$]-terms have the same coefficients.], [Circle],
))

#linebreak() 
ⓓ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "y equals −2 x squared minus 4 x minus 5")[$y = −2 x^(2) − 4 x − 5$]]),
  [Only one variable, #math.equation(block: false, alt: "x")[$x$], is squared.], [Parabola],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

ⓐ #math.equation(block: false, alt: "x squared plus y squared minus 8 x minus 6 y equals 0")[$x^(2) + y^(2) − 8 x − 6 y = 0$] ⓑ #math.equation(block: false, alt: "4 x squared plus 25 y squared equals 100")[$4 x^(2) + 25 y^(2) = 100$] ⓒ #math.equation(block: false, alt: "y equals 6 x squared plus 2 x minus 1")[$y = 6 x^(2) + 2 x − 1$] ⓓ #math.equation(block: false, alt: "16 y squared minus 9 x squared equals 144")[$16 y^(2) − 9 x^(2) = 144$]

#solutionbox[
ⓐ circle ⓑ ellipse ⓒ parabola ⓓ hyperbola
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

ⓐ #math.equation(block: false, alt: "16 x squared plus 9 y squared equals 144")[$16 x^(2) + 9 y^(2) = 144$] ⓑ #math.equation(block: false, alt: "y equals 2 x squared plus 4 x plus 6")[$y = 2 x^(2) + 4 x + 6$] ⓒ #math.equation(block: false, alt: "x squared plus y squared plus 2 x plus 6 y plus 9 equals 0")[$x^(2) + y^(2) + 2 x + 6 y + 9 = 0$] ⓓ #math.equation(block: false, alt: "4 x squared minus 16 y squared equals 64")[$4 x^(2) − 16 y^(2) = 64$]

#solutionbox[
ⓐ ellipse ⓑ parabola ⓒ circle ⓓ hyperbola
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access these online resources for additional instructions and practice with hyperbolas.

- #link("https://openstax.org/l/37ghyperborig")[Graph a Hyperbola with Center at the Origin]
- #link("https://openstax.org/l/37ghyperbnorig")[Graph a Hyperbola with Center not at the Origin]
- #link("https://openstax.org/l/37ghyperbgen")[Graph a Hyperbola in General Form]
- #link("https://openstax.org/l/37conicsgen")[Identifying Conic Sections in General Form]
]


=== Key Concepts

- #strong[Hyperbola:] A #strong[hyperbola] is all points in a plane where the difference of their distances from two fixed points is constant. #linebreak() 
#figure(figph[The figure shows a double napped right circular cone sliced by a plane that is parallel to the vertical axis of the cone forming a hyperbola. The figure is labeled ‘hyperbola’.], alt: "The figure shows a double napped right circular cone sliced by a plane that is parallel to the vertical axis of the cone forming a hyperbola. The figure is labeled ‘hyperbola’.", caption: none)

 #linebreak() 
Each of the fixed points is called a #strong[focus] of the hyperbola. #linebreak() 
The line through the foci, is called the #strong[transverse axis]. #linebreak() 
The two points where the transverse axis intersects the hyperbola are each a #strong[vertex] of the hyperbola. #linebreak() 
The midpoint of the segment joining the foci is called the #strong[center] of the hyperbola. #linebreak() 
The line perpendicular to the transverse axis that passes through the center is called the #strong[conjugate axis]. #linebreak() 
Each piece of the graph is called a #strong[branch] of the hyperbola. #linebreak() 
#figure(figph[The figure shows two graphs of a hyperbola. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci are shown with points that lie on the transverse axis, which is the x-axis. The branches pass through the vertices and open left and right. The y-axis is the conjugate axis. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci lie are shown with points that lie on the transverse axis, which is the y-axis. The branches pass through the vertices and open up and down. The x-axis is the conjugate axis.], alt: "The figure shows two graphs of a hyperbola. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci are shown with points that lie on the transverse axis, which is the x-axis. The branches pass through the vertices and open left and right. The y-axis is the conjugate axis. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci lie are shown with points that lie on the transverse axis, which is the y-axis. The branches pass through the vertices and open up and down. The x-axis is the conjugate axis.", caption: none)

 #linebreak() 
#figure(table(
  columns: 3, align: left, inset: 6pt,
  [Standard Forms of the Equation a Hyperbola with Center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]], [], [],
  [], [#math.equation(block: false, alt: "the fraction x squared over a squared minus the fraction y squared over b squared equals 1")[$frac(x^(2), a^(2)) − frac(y^(2), b^(2)) = 1$]], [#math.equation(block: false, alt: "the fraction y squared over a squared minus the fraction x squared over b squared equals 1")[$frac(y^(2), a^(2)) − frac(x^(2), b^(2)) = 1$]],
  [#strong[Orientation]], [Transverse axis on the #emph[x]-axis. #linebreak() Opens left and right], [Transverse axis on the #emph[y]-axis. #linebreak() Opens up and down],
  [#strong[Vertices]], [#math.equation(block: false, alt: "open parenthesis − a , 0 close parenthesis ,")[$( "−" a , 0 ) ,$] #math.equation(block: false, alt: "open parenthesis a , 0 close parenthesis")[$( a , 0 )$]], [#math.equation(block: false, alt: "open parenthesis 0 , − a close parenthesis ,")[$( 0 , "−" a ) ,$] #math.equation(block: false, alt: "open parenthesis 0 , a close parenthesis")[$( 0 , a )$]],
  [#strong[#emph[x]-intercepts]], [#math.equation(block: false, alt: "open parenthesis − a , 0 close parenthesis ,")[$( "−" a , 0 ) ,$] #math.equation(block: false, alt: "open parenthesis a , 0 close parenthesis")[$( a , 0 )$]], [none],
  [#strong[#emph[y]-intercepts]], [none], [#math.equation(block: false, alt: "open parenthesis 0 , − a close parenthesis")[$( 0 , "−" a )$], #math.equation(block: false, alt: "open parenthesis 0 , a close parenthesis")[$( 0 , a )$]],
  [#strong[Rectangle]], [Use #math.equation(block: false, alt: "open parenthesis ± a , 0 close parenthesis")[$( "±" a , 0 )$] #math.equation(block: false, alt: "open parenthesis 0 , ± b close parenthesis")[$( 0 , "±" b )$]], [Use #math.equation(block: false, alt: "open parenthesis 0 , ± a close parenthesis")[$( 0 , "±" a )$] #math.equation(block: false, alt: "open parenthesis ± b , 0 close parenthesis")[$( "±" b , 0 )$]],
  [#strong[asymptotes]], [#math.equation(block: false, alt: "y equals the fraction b over a x ,")[$y = frac(b, a) x ,$] #math.equation(block: false, alt: "y equals minus the fraction b over a x")[$y = − frac(b, a) x$]], [#math.equation(block: false, alt: "y equals the fraction a over b x ,")[$y = frac(a, b) x ,$] #math.equation(block: false, alt: "y equals minus the fraction a over b x")[$y = − frac(a, b) x$]],
))
- #strong[How to graph a hyperbola centered at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$]]
+ Write the equation in standard form.
+ Determine whether the transverse axis is horizontal or vertical.
+ Find the vertices.
+ Sketch the rectangle centered at the origin intersecting one axis at #math.equation(block: false, alt: "± a")[$"±" a$] and the other at #math.equation(block: false, alt: "± b .")[$"±" b .$]
+ Sketch the asymptotes—the lines through the diagonals of the rectangle.
+ Draw the two branches of the hyperbola.

 #linebreak() 
#figure(table(
  columns: 3, align: left, inset: 6pt,
  [Standard Forms of the Equation a Hyperbola with Center #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]], [], [],
  [], [#math.equation(block: false, alt: "the fraction open parenthesis x minus h close parenthesis squared over a squared minus the fraction open parenthesis y minus k close parenthesis squared over b squared equals 1")[$frac(attach(( x − h ), t: 2), a^(2)) − frac(attach(( y − k ), t: 2), b^(2)) = 1$]], [#math.equation(block: false, alt: "the fraction open parenthesis y minus k close parenthesis squared over a squared minus the fraction open parenthesis x minus h close parenthesis squared over b squared equals 1")[$frac(attach(( y − k ), t: 2), a^(2)) − frac(attach(( x − h ), t: 2), b^(2)) = 1$]],
  [#strong[Orientation]], [Transverse axis is horizontal. #linebreak() Opens left and right], [Transverse axis is vertical. #linebreak() Opens up and down],
  [#strong[Center]], [#math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]], [#math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]],
  [#strong[Vertices]], [#emph[a] units to the left and right of the center], [#emph[a] units above and below the center],
  [#strong[Rectangle]], [Use #emph[a] units left/right of center #linebreak() #emph[b] units above/below the center], [Use #emph[a] units above/below the center #linebreak() #emph[b] units left/right of center],
))
- #strong[How to graph a hyperbola centered at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis .")[$( h , k ) .$]]
+ Write the equation in standard form.
+ Determine whether the transverse axis is horizontal or vertical.
+ Find the center and #math.equation(block: false, alt: "a , b .")[$a , b .$]
+ Sketch the rectangle centered at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$] using #math.equation(block: false, alt: "a , b .")[$a , b .$]
+ Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices.
+ Draw the two branches of the hyperbola.

 #linebreak() 
#figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([Conic], [Characteristics of #math.equation(block: false, alt: "x squared -")[$x^(2) "-"$] and #math.equation(block: false, alt: "y squared -")[$y^(2) "-"$] terms], [Example]),
  [#strong[Parabola]], [Either #math.equation(block: false, alt: "x squared")[$x^(2)$] OR #math.equation(block: false, alt: "y squared .")[$y^(2) .$] Only one variable is squared.], [#math.equation(block: false, alt: "x equals 3 y squared minus 2 y plus 1")[$x = 3 y^(2) − 2 y + 1$]],
  [#strong[Circle]], [#math.equation(block: false, alt: "x squared -")[$x^(2) "-"$] and #math.equation(block: false, alt: "y squared -")[$y^(2) "-"$] terms must have the same coefficients and they must be the same sign as the constant after the = sign], [#math.equation(block: false, alt: "x squared plus y squared equals 49")[$x^(2) + y^(2) = 49$]],
  [#strong[Ellipse]], [#math.equation(block: false, alt: "x squared -")[$x^(2) "-"$] and #math.equation(block: false, alt: "y squared -")[$y^(2) "-"$] terms have the #strong[same] sign, different coefficients], [#math.equation(block: false, alt: "4 x squared plus 25 y squared equals 100")[$4 x^(2) + 25 y^(2) = 100$]],
  [#strong[Hyperbola]], [#math.equation(block: false, alt: "x squared -")[$x^(2) "-"$] and #math.equation(block: false, alt: "y squared -")[$y^(2) "-"$] terms have #strong[different] signs], [#math.equation(block: false, alt: "25 y squared minus 4 x squared equals 100")[$25 y^(2) − 4 x^(2) = 100$]],
))


==== Practice Makes Perfect

#strong[Graph a Hyperbola with Center at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]]

In the following exercises, graph.

#math.equation(block: true, alt: "the fraction x squared over 9 minus the fraction y squared over 4 equals 1")[$frac(x^(2), 9) − frac(y^(2), 4) = 1$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus two-thirds times x, and branches that pass through the vertices (plus or minus 3, 0) and open left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus two-thirds times x, and branches that pass through the vertices (plus or minus 3, 0) and open left and right.", caption: none)
]

#math.equation(block: true, alt: "the fraction x squared over 25 minus the fraction y squared over 9 equals 1")[$frac(x^(2), 25) − frac(y^(2), 9) = 1$]

#math.equation(block: true, alt: "the fraction x squared over 16 minus the fraction y squared over 25 equals 1")[$frac(x^(2), 16) − frac(y^(2), 25) = 1$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus five-fourths times x, and branches that pass through the vertices (plus or minus 4, 0) and open left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus five-fourths times x, and branches that pass through the vertices (plus or minus 4, 0) and open left and right.", caption: none)
]

#math.equation(block: true, alt: "the fraction x squared over 9 minus the fraction y squared over 36 equals 1")[$frac(x^(2), 9) − frac(y^(2), 36) = 1$]

#math.equation(block: true, alt: "the fraction y squared over 25 minus the fraction x squared over 4 equals 1")[$frac(y^(2), 25) − frac(x^(2), 4) = 1$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus five-halves times x, and branches that pass through the vertices (0, plus or minus 5) and open up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus five-halves times x, and branches that pass through the vertices (0, plus or minus 5) and open up and down.", caption: none)
]

#math.equation(block: true, alt: "the fraction y squared over 36 minus the fraction x squared over 16 equals 1")[$frac(y^(2), 36) − frac(x^(2), 16) = 1$]

#math.equation(block: true, alt: "16 y squared minus 9 x squared equals 144")[$16 y^(2) − 9 x^(2) = 144$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus three-fourths times x, and branches that pass through the vertices (0, plus or minus 3) and open up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus three-fourths times x, and branches that pass through the vertices (0, plus or minus 3) and open up and down.", caption: none)
]

#math.equation(block: true, alt: "25 y squared minus 9 x squared equals 225")[$25 y^(2) − 9 x^(2) = 225$]

#math.equation(block: true, alt: "4 y squared minus 9 x squared equals 36")[$4 y^(2) − 9 x^(2) = 36$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus three-halves times x, and branches that pass through the vertices (0, plus or minus 3) and open up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus three-halves times x, and branches that pass through the vertices (0, plus or minus 3) and open up and down.", caption: none)
]

#math.equation(block: true, alt: "16 y squared minus 25 x squared equals 400")[$16 y^(2) − 25 x^(2) = 400$]

#math.equation(block: true, alt: "4 x squared minus 16 y squared equals 64")[$4 x^(2) − 16 y^(2) = 64$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus one-half times x, and branches that pass through the vertices (plus or minus 4, 0) and open left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus one-half times x, and branches that pass through the vertices (plus or minus 4, 0) and open left and right.", caption: none)
]

#math.equation(block: true, alt: "9 x squared minus 4 y squared equals 36")[$9 x^(2) − 4 y^(2) = 36$]

#strong[Graph a Hyperbola with Center at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]]

In the following exercises, graph.

#math.equation(block: true, alt: "the fraction open parenthesis x minus 1 close parenthesis squared over 16 minus the fraction open parenthesis y minus 3 close parenthesis squared over 4 equals 1")[$frac(attach(( x − 1 ), t: 2), 16) − frac(attach(( y − 3 ), t: 2), 4) = 1$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 3) an asymptote that passes through (negative 3, 1) and (5, 5) and an asymptote that passes through (5, 1) and (negative 3, 5), and branches that pass through the vertices (negative 3, 3) and (5, 3) and opens left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 3) an asymptote that passes through (negative 3, 1) and (5, 5) and an asymptote that passes through (5, 1) and (negative 3, 5), and branches that pass through the vertices (negative 3, 3) and (5, 3) and opens left and right.", caption: none)
]

#math.equation(block: true, alt: "the fraction open parenthesis x minus 2 close parenthesis squared over 4 minus the fraction open parenthesis y minus 3 close parenthesis squared over 16 equals 1")[$frac(attach(( x − 2 ), t: 2), 4) − frac(attach(( y − 3 ), t: 2), 16) = 1$]

#math.equation(block: true, alt: "the fraction open parenthesis y minus 4 close parenthesis squared over 9 minus the fraction open parenthesis x minus 2 close parenthesis squared over 25 equals 1")[$frac(attach(( y − 4 ), t: 2), 9) − frac(attach(( x − 2 ), t: 2), 25) = 1$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 3) an asymptote that passes through (negative 3, 1) and (5, 5) and an asymptote that passes through (5, 1) and (negative 3, 5), and branches that pass through the vertices (negative 3, 3) and (5, 3) and opens left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 3) an asymptote that passes through (negative 3, 1) and (5, 5) and an asymptote that passes through (5, 1) and (negative 3, 5), and branches that pass through the vertices (negative 3, 3) and (5, 3) and opens left and right.", caption: none)
]

#math.equation(block: true, alt: "the fraction open parenthesis y minus 1 close parenthesis squared over 25 minus the fraction open parenthesis x minus 4 close parenthesis squared over 16 equals 1")[$frac(attach(( y − 1 ), t: 2), 25) − frac(attach(( x − 4 ), t: 2), 16) = 1$]

#math.equation(block: true, alt: "the fraction open parenthesis y plus 4 close parenthesis squared over 25 minus the fraction open parenthesis x plus 1 close parenthesis squared over 36 equals 1")[$frac(attach(( y + 4 ), t: 2), 25) − frac(attach(( x + 1 ), t: 2), 36) = 1$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, negative 4) an asymptote that passes through (negative 7, 1) and (5, negative 9) and an asymptote that passes through (5, 1) and (negative 7, negative 9), and branches that pass through the vertices (1, 1) and (1, negative 9) and open up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, negative 4) an asymptote that passes through (negative 7, 1) and (5, negative 9) and an asymptote that passes through (5, 1) and (negative 7, negative 9), and branches that pass through the vertices (1, 1) and (1, negative 9) and open up and down.", caption: none)
]

#math.equation(block: true, alt: "the fraction open parenthesis y plus 1 close parenthesis squared over 16 minus the fraction open parenthesis x plus 1 close parenthesis squared over 4 equals 1")[$frac(attach(( y + 1 ), t: 2), 16) − frac(attach(( x + 1 ), t: 2), 4) = 1$]

#math.equation(block: true, alt: "the fraction open parenthesis y minus 4 close parenthesis squared over 16 minus the fraction open parenthesis x plus 1 close parenthesis squared over 25 equals 1")[$frac(attach(( y − 4 ), t: 2), 16) − frac(attach(( x + 1 ), t: 2), 25) = 1$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (negative 1, 4) an asymptote that passes through (4, 8) and (negative 6, 0) and an asymptote that passes through (negative 6, 8) and (4, 0), and branches that pass through the vertices (negative 1, 0) and (negative 1, 8) and open up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (negative 1, 4) an asymptote that passes through (4, 8) and (negative 6, 0) and an asymptote that passes through (negative 6, 8) and (4, 0), and branches that pass through the vertices (negative 1, 0) and (negative 1, 8) and open up and down.", caption: none)
]

#math.equation(block: true, alt: "the fraction open parenthesis y plus 3 close parenthesis squared over 16 minus the fraction open parenthesis x minus 3 close parenthesis squared over 36 equals 1")[$frac(attach(( y + 3 ), t: 2), 16) − frac(attach(( x − 3 ), t: 2), 36) = 1$]

#math.equation(block: true, alt: "the fraction open parenthesis x minus 3 close parenthesis squared over 25 minus the fraction open parenthesis y plus 2 close parenthesis squared over 9 equals 1")[$frac(attach(( x − 3 ), t: 2), 25) − frac(attach(( y + 2 ), t: 2), 9) = 1$]

#solutionbox[
#figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (3, negative 2) an asymptote that passes through (8, 1) and (negative 2, negative 5) and an asymptote that passes through (negative 2, negative 1) and (8, negative 5), and branches that pass through the vertices (negative 2, negative 2) and (8, negative 2) and opens left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (3, negative 2) an asymptote that passes through (8, 1) and (negative 2, negative 5) and an asymptote that passes through (negative 2, negative 1) and (8, negative 5), and branches that pass through the vertices (negative 2, negative 2) and (8, negative 2) and opens left and right.", caption: none)
]

#math.equation(block: true, alt: "the fraction open parenthesis x plus 2 close parenthesis squared over 4 minus the fraction open parenthesis y minus 1 close parenthesis squared over 9 equals 1")[$frac(attach(( x + 2 ), t: 2), 4) − frac(attach(( y − 1 ), t: 2), 9) = 1$]

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

#math.equation(block: true, alt: "9 x squared minus 4 y squared minus 18 x plus 8 y minus 31 equals 0")[$9 x^(2) − 4 y^(2) − 18 x + 8 y − 31 = 0$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "the fraction open parenthesis x minus 1 close parenthesis squared over 4 minus the fraction open parenthesis y minus 1 close parenthesis squared over 9 equals 1")[$frac(attach(( x − 1 ), t: 2), 4) − frac(attach(( y − 1 ), t: 2), 9) = 1$] #linebreak() 
ⓑ  #linebreak() #figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 1) an asymptote that passes through (3, 4) and (negative 1, negative 2) and an asymptote that passes through (negative 1, 4) and (3, negative 2), and branches that pass through the vertices (negative 1, 1) and (3, 1) and opens left and right.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 1) an asymptote that passes through (3, 4) and (negative 1, negative 2) and an asymptote that passes through (negative 1, 4) and (3, negative 2), and branches that pass through the vertices (negative 1, 1) and (3, 1) and opens left and right.", caption: none)
]

#math.equation(block: true, alt: "16 x squared minus 4 y squared plus 64 x minus 24 y minus 36 equals 0")[$16 x^(2) − 4 y^(2) + 64 x − 24 y − 36 = 0$]

#math.equation(block: true, alt: "y squared minus x squared minus 4 y plus 2 x minus 6 equals 0")[$y^(2) − x^(2) − 4 y + 2 x − 6 = 0$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "the fraction open parenthesis y minus 2 close parenthesis squared over 9 minus the fraction open parenthesis x minus 1 close parenthesis squared over 9 equals 1")[$frac(attach(( y − 2 ), t: 2), 9) − frac(attach(( x − 1 ), t: 2), 9) = 1$] #linebreak() 
ⓑ  #linebreak() #figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 2) an asymptote that passes through (4, 5) and (negative 2, negative 1) and an asymptote that passes through (negative 2, 5) and (4, negative 1), and branches that pass through the vertices (1, 5) and (1, negative 1) and open up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 2) an asymptote that passes through (4, 5) and (negative 2, negative 1) and an asymptote that passes through (negative 2, 5) and (4, negative 1), and branches that pass through the vertices (1, 5) and (1, negative 1) and open up and down.", caption: none)
]

#math.equation(block: true, alt: "4 y squared minus 16 x squared minus 24 y plus 96 x minus 172 equals 0")[$4 y^(2) − 16 x^(2) − 24 y + 96 x − 172 = 0$]

#math.equation(block: true, alt: "9 y squared minus x squared plus 18 y minus 4 x minus 4 equals 0")[$9 y^(2) − x^(2) + 18 y − 4 x − 4 = 0$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "the fraction open parenthesis y plus 1 close parenthesis squared over 1 minus the fraction open parenthesis x plus 2 close parenthesis squared over 9 equals 1")[$frac(attach(( y + 1 ), t: 2), 1) − frac(attach(( x + 2 ), t: 2), 9) = 1$] #linebreak() 
ⓑ  #linebreak() #figure(figph[The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (negative 2, negative 1) an asymptote that passes through (1, 0) and (negative 5, negative 2) and an asymptote that passes through (3, 0) and (1, negative 2), and branches that pass through the vertices (negative 2, 0) and (negative 2, negative 2) and open up and down.], alt: "The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (negative 2, negative 1) an asymptote that passes through (1, 0) and (negative 5, negative 2) and an asymptote that passes through (3, 0) and (1, negative 2), and branches that pass through the vertices (negative 2, 0) and (negative 2, negative 2) and open up and down.", caption: none)
]

#strong[Identify the Graph of each Equation as a Circle, Parabola, Ellipse, or Hyperbola]

In the following exercises, identify the type of graph.

ⓐ #math.equation(block: false, alt: "x equals − y squared minus 2 y plus 3")[$x = "−" y^(2) − 2 y + 3$] ⓑ #math.equation(block: false, alt: "9 y squared minus x squared plus 18 y minus 4 x minus 4 equals 0")[$9 y^(2) − x^(2) + 18 y − 4 x − 4 = 0$] ⓒ #math.equation(block: false, alt: "9 x squared plus 25 y squared equals 225")[$9 x^(2) + 25 y^(2) = 225$] ⓓ #math.equation(block: false, alt: "x squared plus y squared minus 4 x plus 10 y minus 7 equals 0")[$x^(2) + y^(2) − 4 x + 10 y − 7 = 0$]

ⓐ #math.equation(block: false, alt: "x equals −2 y squared minus 12 y minus 16")[$x = −2 y^(2) − 12 y − 16$] ⓑ #math.equation(block: false, alt: "x squared plus y squared equals 9")[$x^(2) + y^(2) = 9$] ⓒ #math.equation(block: false, alt: "16 x squared minus 4 y squared plus 64 x minus 24 y minus 36 equals 0")[$16 x^(2) − 4 y^(2) + 64 x − 24 y − 36 = 0$] ⓓ #math.equation(block: false, alt: "16 x squared plus 36 y squared equals 576")[$16 x^(2) + 36 y^(2) = 576$]

#solutionbox[
ⓐ parabola ⓑ circle ⓒ hyperbola ⓓ ellipse
]

#strong[Mixed Practice]

In the following exercises, graph each equation.

#math.equation(block: true, alt: "the fraction open parenthesis y minus 3 close parenthesis squared over 9 minus the fraction open parenthesis x plus 2 close parenthesis squared over 16 equals 1")[$frac(attach(( y − 3 ), t: 2), 9) − frac(attach(( x + 2 ), t: 2), 16) = 1$]

#math.equation(block: true, alt: "x squared plus y squared minus 4 x plus 10 y minus 7 equals 0")[$x^(2) + y^(2) − 4 x + 10 y − 7 = 0$]

#solutionbox[
#figure(figph[The graph shows the x y coordinate plane with a circle whose center is (2, negative 5) and whose radius is 6 units.], alt: "The graph shows the x y coordinate plane with a circle whose center is (2, negative 5) and whose radius is 6 units.", caption: none)
]

#math.equation(block: true, alt: "y equals open parenthesis x minus 1 close parenthesis squared plus 2")[$y = attach(( x − 1 ), t: 2) + 2$]

#math.equation(block: true, alt: "the fraction x squared over 9 plus the fraction y squared over 25 equals 1")[$frac(x^(2), 9) + frac(y^(2), 25) = 1$]

#solutionbox[
#figure(figph[The graph shows the x y coordinate plane with an ellipse whose major axis is vertical, vertices are (0, plus or minus 5) and co-vertices are (plus or minus 3, 0).], alt: "The graph shows the x y coordinate plane with an ellipse whose major axis is vertical, vertices are (0, plus or minus 5) and co-vertices are (plus or minus 3, 0).", caption: none)
]

#math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis squared plus open parenthesis y minus 5 close parenthesis squared equals 4")[$attach(( x + 2 ), t: 2) + attach(( y − 5 ), t: 2) = 4$]

#math.equation(block: true, alt: "y squared minus x squared minus 4 y plus 2 x minus 6 equals 0")[$y^(2) − x^(2) − 4 y + 2 x − 6 = 0$]

#solutionbox[
#figure(figph[The graph shows the x y coordinate plane with the center (1, 2) an asymptote that passes through (negative 2, 5) and (5, negative 1) and an asymptote that passes through (4, 5) and (2, 0), and branches that pass through the vertices (1, 5) and (negative 2, negative 1) and open up and down.], alt: "The graph shows the x y coordinate plane with the center (1, 2) an asymptote that passes through (negative 2, 5) and (5, negative 1) and an asymptote that passes through (4, 5) and (2, 0), and branches that pass through the vertices (1, 5) and (negative 2, negative 1) and open up and down.", caption: none)
]

#math.equation(block: true, alt: "x equals − y squared minus 2 y plus 3")[$x = "−" y^(2) − 2 y + 3$]

#math.equation(block: true, alt: "16 x squared plus 9 y squared equals 144")[$16 x^(2) + 9 y^(2) = 144$]

#solutionbox[
#figure(figph[The graph shows the x y coordinate plane with an ellipse whose major axis is vertical, vertices are (0, plus or minus 4) and co-vertices are (plus or minus 3, 0).], alt: "The graph shows the x y coordinate plane with an ellipse whose major axis is vertical, vertices are (0, plus or minus 4) and co-vertices are (plus or minus 3, 0).", caption: none)
]


==== Writing Exercises

In your own words, define a hyperbola and write the equation of a hyperbola centered at the origin in standard form. Draw a sketch of the hyperbola labeling the center, vertices, and asymptotes.

Explain in your own words how to create and use the rectangle that helps graph a hyperbola.

#solutionbox[
Answers will vary.
]

Compare and contrast the graphs of the equations #math.equation(block: false, alt: "the fraction x squared over 4 minus the fraction y squared over 9 equals 1")[$frac(x^(2), 4) − frac(y^(2), 9) = 1$] and #math.equation(block: false, alt: "the fraction y squared over 9 minus the fraction x squared over 4 equals 1 .")[$frac(y^(2), 9) − frac(x^(2), 4) = 1 .$]

Explain in your own words, how to distinguish the equation of an ellipse with the equation of a hyperbola.

#solutionbox[
Answers will vary.
]


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This table has four columns and four rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was graph a hyperbola with center at (0, 0). In row 3, the I can was graph a hyperbola with a center at (h, k). In row 4, the I can was identify conic sections by their equations.], alt: "This table has four columns and four rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was graph a hyperbola with center at (0, 0). In row 3, the I can was graph a hyperbola with a center at (h, k). In row 4, the I can was identify conic sections by their equations.", caption: none)

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?