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== 11.3#h(0.6em)Ellipses

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

Graph #math.equation(block: false, alt: "y equals open parenthesis x minus 1 close parenthesis squared minus 2")[$y = attach(( x − 1 ), t: 2) − 2$] using transformations. #linebreak() 
If you missed this problem, review .

#solutionbox[
#figure(figph[A graph displays a blue upward-opening parabola on a Cartesian coordinate plane. The curve has its vertex at approximately (1, -2) and crosses the x-axis at roughly x=0 and x=2.5. Grid lines are visible.], alt: "A graph displays a blue upward-opening parabola on a Cartesian coordinate plane. The curve has its vertex at approximately (1, -2) and crosses the x-axis at roughly x=0 and x=2.5. Grid lines are visible.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Complete the square: #math.equation(block: false, alt: "x squared minus 8 x equals 8 .")[$x^(2) − 8 x = 8 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x minus 4 squared equals 8")[$attach(x − 4, t: 2) = 8$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write in standard form. #math.equation(block: false, alt: "y equals 2 x squared minus 12 x plus 14")[$y = 2 x^(2) − 12 x + 14$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "y equals 2 x minus 3 squared minus 4")[$y = 2 attach(x − 3, t: 2) − 4$]
]
]


=== Graph an Ellipse with Center at the Origin

The next conic section we will look at is an #strong[ellipse]. We define an ellipse as all points in a plane where the sum of the distances from two fixed points is constant. Each of the given points is called a #strong[focus] of the ellipse.

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

An #strong[ellipse] is all points in a plane where the sum of the distances from two fixed points is constant. Each of the fixed points is called a #strong[focus] of the ellipse.

#figure(figph[This figure shows a double cone intersected by a plane to form an ellipse.], alt: "This figure shows a double cone intersected by a plane to form an ellipse.", caption: none)
]

We can draw an ellipse by taking some fixed length of flexible string and attaching the ends to two thumbtacks. We use a pen to pull the string taut and rotate it around the two thumbtacks. The figure that results is an ellipse.

#figure(figph[This figure shows a pen attached to two strings, the other ends of which are attached to two thumbtacks. The strings are pulled taut and the pen is rotated to draw an ellipse. The thumbtacks are labeled F subscript 1 and F subscript 2.], alt: "This figure shows a pen attached to two strings, the other ends of which are attached to two thumbtacks. The strings are pulled taut and the pen is rotated to draw an ellipse. The thumbtacks are labeled F subscript 1 and F subscript 2.", caption: none)

A line drawn through the foci intersect the ellipse in two points. Each point is called a #strong[vertex] of the ellipse. The segment connecting the vertices is called the #strong[major axis]. The midpoint of the segment is called the #strong[center] of the ellipse. A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the #strong[minor axis].

#figure(figph[This figure shows two ellipses. In each, two points within the ellipse are labeled foci. A line drawn through the foci intersects the ellipse in two points. Each point is labeled a vertex. In the figure on the left, the segment connecting the vertices is called the major axis. A segment perpendicular to the major axis that passes through its midpoint and intersects the ellipse in two points is labeled minor axis. The minor axis is shorter than the minor axis. In the figure on the right, the segment through the foci, connecting the vertices is longer and is labeled major axis. Its midpoint is labeled center.], alt: "This figure shows two ellipses. In each, two points within the ellipse are labeled foci. A line drawn through the foci intersects the ellipse in two points. Each point is labeled a vertex. In the figure on the left, the segment connecting the vertices is called the major axis. A segment perpendicular to the major axis that passes through its midpoint and intersects the ellipse in two points is labeled minor axis. The minor axis is shorter than the minor axis. In the figure on the right, the segment through the foci, connecting the vertices is longer and is labeled major axis. Its midpoint is labeled center.", caption: none)

We mentioned earlier that our goal is to connect the geometry of a conic with algebra. Placing the ellipse on a rectangular coordinate system gives us that opportunity. In the figure, we placed the ellipse 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 on the left shows an ellipse with its center at the origin of the coordinate axes and its foci at points minus (c, 0) and (c, 0). A segment connects (negative c, 0) to a point (x, y) on the ellipse. The segment is labeled d subscript 1. Another segment, labeled d subscript 2 connects (c, 0) to (x, y). The figure on the right shows an ellipse with center at the origin, foci (negative c, 0) and (c, 0) and vertices (negative a, 0) and (a, 0). The point where the ellipse intersects the y axis is labeled (0, b). The segments connecting (0, 0) to (c, 0), (c, 0) to (0, b) and (0, b) to (0, 0) form a tight angled triangle with sides c, a and b respectively. The equation is a squared equals b squared plus c squared.], alt: "The figure on the left shows an ellipse with its center at the origin of the coordinate axes and its foci at points minus (c, 0) and (c, 0). A segment connects (negative c, 0) to a point (x, y) on the ellipse. The segment is labeled d subscript 1. Another segment, labeled d subscript 2 connects (c, 0) to (x, y). The figure on the right shows an ellipse with center at the origin, foci (negative c, 0) and (c, 0) and vertices (negative a, 0) and (a, 0). The point where the ellipse intersects the y axis is labeled (0, b). The segments connecting (0, 0) to (c, 0), (c, 0) to (0, b) and (0, b) to (0, 0) form a tight angled triangle with sides c, a and b respectively. The equation is a squared equals b squared plus c squared.", caption: none)

The definition states the sum 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 plus 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 plus 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: "Use the distance formula to find d sub 1 , d sub 2 ., d sub 1, plus, d sub 2, equals, 2 a; the square root of open parenthesis x minus open parenthesis − c close parenthesis close parenthesis squared plus open parenthesis y minus 0 close parenthesis squared, plus, the square root of open parenthesis x minus c close parenthesis squared plus open parenthesis y minus 0 close parenthesis squared, equals, 2 a; After eliminating radicals and simplifying,; we get:, the fraction x squared over a squared plus the fraction y squared over a squared minus c squared equals 1 ; To simplify the equation of the ellipse, we; let a squared minus c squared equals b squared .; So, the equation of an ellipse centered at the; origin in standard form is:, the fraction x squared over a squared plus the fraction y")[$"Use the distance formula to find" #h(0.2em) d_(1) , d_(2) . &  &  & d_(1) & + & d_(2) & = & 2 a \  \  \ sqrt(attach(( x − ( "−" c ) ), t: 2) + attach(( y − 0 ), t: 2)) & + & sqrt(attach(( x − c ), t: 2) + attach(( y − 0 ), t: 2)) & = & 2 a \  \  \ "After eliminating radicals and simplifying," \ "we get:" &  &  & frac(x^(2), a^(2)) + frac(y^(2), a^(2) − c^(2)) #h(0.5em) = #h(0.5em) 1 #h(0.6em) \ "To simplify the equation of the ellipse, we" \ "let" #h(0.2em) a^(2) − c^(2) = b^(2) . &  &  &  \ "So, the equation of an ellipse centered at the" \ "origin in standard form is:" &  &  & frac(x^(2), a^(2)) + frac(y^(2), b^(2)) #h(0.5em) = #h(0.5em) 1 #h(0.6em)$]To graph the ellipse, it will be helpful to know the intercepts. We will find the #emph[x]-intercepts and #emph[y]-intercepts using the formula.

#math.equation(block: true, alt: "y -intercepts; Let x equals 0 ., the fraction x squared over a squared plus the fraction y squared over b squared, equals, 1; the fraction 0 squared over a squared plus the fraction y squared over b squared, equals, 1; the fraction y squared over b squared, equals, 1; y squared, equals, b squared; y, equals, plus or minus b, x -intercepts; Let y equals 0 ., the fraction x squared over a squared plus the fraction y squared over b squared, equals, 1; the fraction x squared over a squared plus the fraction 0 squared over b squared, equals, 1; the fraction x squared over a squared, equals, 1; x squared, equals, a squared; x, equals, plus or minus a; The y -intercepts are open parenthesis 0 , b close parenthesis and open parenthesis 0 , − b close parenthesis ., The x -intercepts are open parent")[$&  &  & #h(0.5em) "y" "-intercepts" \ "Let" #h(0.2em) x = 0 . &  &  & frac(x^(2), a^(2)) + frac(y^(2), b^(2)) & = & 1 \ frac(0^(2), a^(2)) + frac(y^(2), b^(2)) & = & 1 \ frac(y^(2), b^(2)) & = & 1 \ y^(2) & = & b^(2) \ y & = & ± b &  &  &  &  &  & #h(0.5em) "x" "-intercepts" \ "Let" #h(0.2em) y = 0 . &  &  & frac(x^(2), a^(2)) + frac(y^(2), b^(2)) & = & 1 \ frac(x^(2), a^(2)) + frac(0^(2), b^(2)) & = & 1 \ frac(x^(2), a^(2)) & = & 1 \ x^(2) & = & a^(2) \ x & = & ± a \ "The" #h(0.2em) y "-intercepts are" #h(0.2em) ( 0 , b ) #h(0.2em) "and" #h(0.2em) ( 0 , "−" b ) . &  &  & "The" #h(0.2em) x "-intercepts are" #h(0.2em) ( a , 0 ) #h(0.2em) "and" #h(0.2em) ( "−" a , 0 ) .$]#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Standard Form of the Equation an Ellipse with Center #math.equation(block: false, alt: "open parenthesis 0 , ​ ​ 0 close parenthesis")[$( 0 , "​" "​" 0 )$]]

The standard form of the equation of an ellipse 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 plus the fraction y squared over b squared equals 1")[$frac(x^(2), a^(2)) + frac(y^(2), b^(2)) = 1$]The #emph[x]-intercepts are #math.equation(block: false, alt: "open parenthesis a , 0 close parenthesis")[$( a , 0 )$] and #math.equation(block: false, alt: "open parenthesis − a , 0 close parenthesis .")[$( "−" a , 0 ) .$]

The #emph[y]-intercepts are #math.equation(block: false, alt: "open parenthesis 0 , b close parenthesis")[$( 0 , b )$] and #math.equation(block: false, alt: "open parenthesis 0 , − b close parenthesis .")[$( 0 , "−" b ) .$]

#figure(figph[Two figures show ellipses with their centers on the origin of the coordinate axes. They intersect the x axis at points (negative a, 0) and (a, 0) and the y axis at points (0, b) and (0, negative b). In the figure on the left the major axis of the ellipse is along the x axis and in the figure on the right, it is along the y axis.], alt: "Two figures show ellipses with their centers on the origin of the coordinate axes. They intersect the x axis at points (negative a, 0) and (a, 0) and the y axis at points (0, b) and (0, negative b). In the figure on the left the major axis of the ellipse is along the x axis and in the figure on the right, it is along the y axis.", caption: none)
]

Notice that when the major axis is horizontal, the value of #emph[a] will be greater than the value of #emph[b] and when the major axis is vertical, the value of #emph[b] will be greater than the value of #emph[a]. We will use this information to graph an #strong[ellipse] that is centered at the origin.

#figure(table(
  columns: 3, align: left, inset: 6pt,
  [Ellipse 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 plus 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: "a greater than b")[$a > b$]], [#math.equation(block: false, alt: "b greater than a")[$b > a$]],
  [#strong[Major axis]], [on the #emph[x]- axis.], [on the #emph[y]-axis.],
  [#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")[$( a , 0 )$]], [],
  [#strong[#emph[y]-intercepts]], [#math.equation(block: false, alt: "open parenthesis 0 , − b close parenthesis ,")[$( 0 , "−" b ) ,$]#math.equation(block: false, alt: "open parenthesis 0 , b close parenthesis")[$( 0 , b )$]], [],
))

#examplebox("Example 1")[How to Graph an Ellipse with Center (0, 0)][
Graph: #math.equation(block: false, alt: "the fraction x squared over 4 plus the fraction y squared over 9 equals 1 .")[$frac(x^(2), 4) + frac(y^(2), 9) = 1 .$]

#solutionbox[
#figure(figph[Step 1. Write the equation in standard form. It is in standard form x squared upon 6 plus y squared upon 9 equals 1.], alt: "Step 1. Write the equation in standard form. It is in standard form x squared upon 6 plus y squared upon 9 equals 1.", caption: none)


#figure(figph[Step 2. Determine whether the major axis is horizontal or vertical. Since 9 is greater than 4 and 9 is in the y squared term, the major axis is vertical.], alt: "Step 2. Determine whether the major axis is horizontal or vertical. Since 9 is greater than 4 and 9 is in the y squared term, the major axis is vertical.", caption: none)


#figure(figph[Step 3. Find the endpoints of the major axis. The endpoints will be the y-intercepts. Since b squared is 9, b is plus or minus 3. The endpoints of the major axis are (0, 3) and (0, negative 3).], alt: "Step 3. Find the endpoints of the major axis. The endpoints will be the y-intercepts. Since b squared is 9, b is plus or minus 3. The endpoints of the major axis are (0, 3) and (0, negative 3).", caption: none)


#figure(figph[Step 4. Find the endpoints of the minor axis. The endpoints will be the x-intercepts. Since a squared is 4, a is plus or minus 2. The endpoints of the minor axis are (2, 0) and (negative 2, 0).], alt: "Step 4. Find the endpoints of the minor axis. The endpoints will be the x-intercepts. Since a squared is 4, a is plus or minus 2. The endpoints of the minor axis are (2, 0) and (negative 2, 0).", caption: none)


#figure(figph[Step 5. Sketch the ellipse using the x and y intercepts. The graph shows an ellipse with center at (0, 0) and foci at (0, 3), (0, negative 3), (negative 2, 0), and (2, 0).], alt: "Step 5. Sketch the ellipse using the x and y intercepts. The graph shows an ellipse with center at (0, 0) and foci at (0, 3), (0, negative 3), (negative 2, 0), and (2, 0).", caption: none)
]
]

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

#solutionbox[
#figure(figph[This graph shows an ellipse with x intercepts (negative 2, 0) and (2, 0) and y intercepts (0, 4) and (0, negative 4).], alt: "This graph shows an ellipse with x intercepts (negative 2, 0) and (2, 0) and y intercepts (0, 4) and (0, negative 4).", caption: none)
]
]

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

#solutionbox[
#figure(figph[This graph shows an ellipse with x intercepts (negative 3, 0) and (3, 0) and y intercepts (0, 4) and (0, negative 4).], alt: "This graph shows an ellipse with x intercepts (negative 3, 0) and (3, 0) and y intercepts (0, 4) and (0, negative 4).", caption: none)
]
]

We summarize the steps for reference.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[How to Graph an Ellipse with Center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$]]

+ Write the equation in standard form.
+ Determine whether the major axis is horizontal or vertical.
+ Find the endpoints of the major axis.
+ Find the endpoints of the minor axis
+ Sketch the ellipse.
]

Sometimes our equation will first need to be put in standard form.

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

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([We recognize this as the equation of an #linebreak() ellipse since both the #emph[x] and #emph[y] terms are #linebreak() squared and have different coefficients.], [#math.equation(block: false, alt: "x squared plus 4 y squared equals 16")[$#h(0.9em) x^(2) + 4 y^(2) = 16$]]),
  [To get the equation in standard form, divide #linebreak() both sides by 16 so that the equation is equal #linebreak() to 1.], [#math.equation(block: false, alt: "the fraction x squared over 16 plus the fraction 4 y squared over 16 equals the fraction 16 over 16")[$#h(0.61em) frac(x^(2), 16) + frac(4 y^(2), 16) = frac(16, 16)$]],
  [Simplify.], [#math.equation(block: false, alt: "the fraction x squared over 16 plus the fraction y squared over 4 equals 1")[$#h(0.67em) frac(x^(2), 16) + frac(y^(2), 4) = 1$]],
  [The equation is in standard form. #linebreak() The ellipse is centered at the origin.], [The center is #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$]],
  [Since #math.equation(block: false, alt: "16 greater than 4")[$16 > 4$] and 16 is in the #math.equation(block: false, alt: "x squared")[$x^(2)$] term, #linebreak() the major axis is horizontal.], [],
  [#math.equation(block: false, alt: "a squared equals 16 , a equals plus or minus 4")[$a^(2) = 16 , a = ± 4$] #linebreak()   #math.equation(block: false, alt: "b squared equals 4 , b equals plus or minus 2")[$b^(2) = 4 , #h(0.55em) b = ± 2$]], [The vertices are #math.equation(block: false, alt: "open parenthesis 4 , 0 close parenthesis , open parenthesis −4 , 0 close parenthesis .")[$( 4 , 0 ) , ( −4 , 0 ) .$] #linebreak() The endpoints of the minor axis are #linebreak() #math.equation(block: false, alt: "open parenthesis 0 , 2 close parenthesis , open parenthesis 0 , −2 close parenthesis .")[$( 0 , 2 ) , ( 0 , −2 ) .$]],
  [Sketch the ellipse.], [#figure(figph[A graph displays an ellipse centered at the origin (0,0) on a Cartesian coordinate system. It passes through points (-4,0), (4,0), (0,2), and (0,-2).], alt: "A graph displays an ellipse centered at the origin (0,0) on a Cartesian coordinate system. It passes through points (-4,0), (4,0), (0,2), and (0,-2).", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "9 x squared plus 16 y squared equals 144 .")[$9 x^(2) + 16 y^(2) = 144 .$]

#solutionbox[
#figure(figph[This graph shows an ellipse with x intercepts (negative 4, 0) and (4, 0) and y intercepts (0, 3) and (0, negative 3).], alt: "This graph shows an ellipse with x intercepts (negative 4, 0) and (4, 0) and y intercepts (0, 3) and (0, negative 3).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "16 x squared plus 25 y squared equals 400 .")[$16 x^(2) + 25 y^(2) = 400 .$]

#solutionbox[
#figure(figph[This graph shows an ellipse with x intercepts (negative 5, 0) and (5, 0) and y intercepts (0, 4) and (0, negative 4).], alt: "This graph shows an ellipse with x intercepts (negative 5, 0) and (5, 0) and y intercepts (0, 4) and (0, negative 4).", caption: none)
]
]


=== Find the Equation of an Ellipse with Center at the Origin

If we are given the graph of an #strong[ellipse], we can find the equation of the ellipse.

#examplebox("Example 3")[][
Find the equation of the ellipse shown.

#figure(figph[This graph shows an ellipse with x intercepts (negative 4, 0) and (4, 0) and y intercepts (0, 3) and (0, negative 3).], alt: "This graph shows an ellipse with x intercepts (negative 4, 0) and (4, 0) and y intercepts (0, 3) and (0, negative 3).", caption: none)

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([We recognize this as an ellipse that is centered at the origin.], [#math.equation(block: false, alt: "the fraction x squared over a squared plus the fraction y squared over b squared equals 1")[$frac(x^(2), a^(2)) + frac(y^(2), b^(2)) = 1$]]),
  [Since the major axis is horizontal and the distance from the center to the vertex is 4, we know #math.equation(block: false, alt: "a equals 4")[$a = 4$] and so #math.equation(block: false, alt: "a squared equals 16")[$a^(2) = 16$].], [#math.equation(block: false, alt: "the fraction x squared over 16 plus the fraction y squared over b squared equals 1")[$frac(x^(2), 16) + frac(y^(2), b^(2)) = 1$]],
  [The minor axis is vertical and the distance from the center to the ellipse is 3, we know #math.equation(block: false, alt: "b equals 3")[$b = 3$] and so #math.equation(block: false, alt: "b squared equals 9")[$b^(2) = 9$].], [#math.equation(block: false, alt: "the fraction x squared over 16 plus the fraction y squared over 9 equals 1")[$frac(x^(2), 16) + frac(y^(2), 9) = 1$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the equation of the ellipse shown.

#figure(figph[This graph shows an ellipse with x intercepts (negative 2, 0) and (2, 0) and y intercepts (0, 5) and (0, negative 5).], alt: "This graph shows an ellipse with x intercepts (negative 2, 0) and (2, 0) and y intercepts (0, 5) and (0, negative 5).", caption: none)

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the equation of the ellipse shown.

#figure(figph[This graph shows an ellipse with x intercepts (negative 3, 0) and (3, 0) and y intercepts (0, 2) and (0, negative 2).], alt: "This graph shows an ellipse with x intercepts (negative 3, 0) and (3, 0) and y intercepts (0, 2) and (0, negative 2).", caption: none)

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


=== Graph an Ellipse with Center Not at the Origin

The ellipses we have looked at so far have all been centered at the origin. We will now look at ellipses whose center is #math.equation(block: false, alt: "open parenthesis h , k close parenthesis .")[$( h , k ) .$]

The equation is #math.equation(block: false, alt: "the fraction open parenthesis x minus h close parenthesis squared over a squared plus the fraction open parenthesis y minus k close parenthesis squared over b squared equals 1")[$frac(attach(( x − h ), t: 2), a^(2)) + frac(attach(( y − k ), t: 2), b^(2)) = 1$] and when #math.equation(block: false, alt: "a greater than b ,")[$a > b ,$] the major axis is horizontal so the distance from the center to the vertex is #emph[a]. When #math.equation(block: false, alt: "b greater than a ,")[$b > a ,$] the major axis is vertical so the distance from the center to the vertex is #emph[b].

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

The standard form of the equation of an ellipse with center #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] is

#math.equation(block: true, alt: "the fraction open parenthesis x minus h close parenthesis squared over a squared plus the fraction open parenthesis y minus k close parenthesis squared over b squared equals 1")[$frac(attach(( x − h ), t: 2), a^(2)) + frac(attach(( y − k ), t: 2), b^(2)) = 1$]When #math.equation(block: false, alt: "a greater than b ,")[$a > b ,$] the major axis is horizontal so the distance from the center to the vertex is #emph[a].

When #math.equation(block: false, alt: "b greater than a ,")[$b > a ,$] the major axis is vertical so the distance from the center to the vertex is #emph[b].
]

#examplebox("Example 4")[][
Graph: #math.equation(block: false, alt: "the fraction open parenthesis x minus 3 close parenthesis squared over 9 plus the fraction open parenthesis y minus 1 close parenthesis squared over 4 equals 1 .")[$frac(attach(( x − 3 ), t: 2), 9) + frac(attach(( y − 1 ), t: 2), 4) = 1 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([The equation is in standard form, #linebreak() #math.equation(block: false, alt: "the fraction open parenthesis x minus h close parenthesis squared over a squared plus 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 x minus 3 close parenthesis squared over 9 plus the fraction open parenthesis y minus 1 close parenthesis squared over 4 equals 1")[$frac(attach(( x − 3 ), t: 2), 9) + frac(attach(( y − 1 ), t: 2), 4) = 1$]]),
  [The ellipse is centered at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis .")[$( h , k ) .$]], [The center is #math.equation(block: false, alt: "open parenthesis 3 , 1 close parenthesis .")[$( 3 , 1 ) .$]],
  [Since #math.equation(block: false, alt: "9 greater than 4")[$9 > 4$] and 9 is in the #math.equation(block: false, alt: "x squared")[$x^(2)$] term, #linebreak() the major axis is horizontal.], [],
  [#math.equation(block: false, alt: "a squared equals 9 , a equals plus or minus 3")[$a^(2) = 9 , #h(0.25em) a = ± 3$] #linebreak()   #math.equation(block: false, alt: "b squared equals 4 , b equals plus or minus 2")[$b^(2) = 4 , #h(0.25em) b = ± 2$]], [The distance from the center to the vertices is 3. #linebreak() The distance from the center to the endpoints of the #linebreak() minor axis is 2.],
  [Sketch the ellipse.], [#figure(figph[An ellipse centered at (3,1) is plotted on a Cartesian coordinate plane, showing its center, vertices at (3,3) and (3,-1), and co-vertices at (0,1) and (6,1).], alt: "An ellipse centered at (3,1) is plotted on a Cartesian coordinate plane, showing its center, vertices at (3,3) and (3,-1), and co-vertices at (0,1) and (6,1).", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph: #math.equation(block: false, alt: "the fraction open parenthesis x plus 3 close parenthesis squared over 4 plus the fraction open parenthesis y minus 5 close parenthesis squared over 16 equals 1 .")[$frac(attach(( x + 3 ), t: 2), 4) + frac(attach(( y − 5 ), t: 2), 16) = 1 .$]

#solutionbox[
#figure(figph[This graph shows an ellipse with center at (negative 3, 5), vertices at (negative 3, 9) and (negative 3, 1) and endpoints of minor axis at (negative 5, 5) and (negative 1, 5).], alt: "This graph shows an ellipse with center at (negative 3, 5), vertices at (negative 3, 9) and (negative 3, 1) and endpoints of minor axis at (negative 5, 5) and (negative 1, 5).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph: #math.equation(block: false, alt: "the fraction open parenthesis x minus 1 close parenthesis squared over 25 plus the fraction open parenthesis y plus 3 close parenthesis squared over 16 equals 1 .")[$frac(attach(( x − 1 ), t: 2), 25) + frac(attach(( y + 3 ), t: 2), 16) = 1 .$]

#solutionbox[
#figure(figph[This graph shows an ellipse with center at 1, negative 3, vertices at (negative 4, negative 3) and (6, negative 3) and endpoints of minor axis at 1, 1) and (negative 1, negative 7).], alt: "This graph shows an ellipse with center at 1, negative 3, vertices at (negative 4, negative 3) and (6, negative 3) and endpoints of minor axis at 1, 1) and (negative 1, negative 7).", caption: none)
]
]

If we look at the equations of #math.equation(block: false, alt: "the fraction x squared over 9 plus the fraction y squared over 4 equals 1")[$frac(x^(2), 9) + frac(y^(2), 4) = 1$] and #math.equation(block: false, alt: "the fraction open parenthesis x minus 3 close parenthesis squared over 9 plus the fraction open parenthesis y minus 1 close parenthesis squared over 4 equals 1 ,")[$frac(attach(( x − 3 ), t: 2), 9) + frac(attach(( y − 1 ), t: 2), 4) = 1 ,$] we see that they are both ellipses with #math.equation(block: false, alt: "a equals 3")[$a = 3$] and #math.equation(block: false, alt: "b equals 2 .")[$b = 2 .$] So they will have the same size and shape. They are different in that they do not have the same center.

#figure(figph[The equation in the first figure is x squared upon 9 plus y squared upon 4 equals 1. Here, a is 3 and b is 2. The ellipse is graphed with center at (0, 0). The equation on the right is open parentheses x minus 3 close parentheses squared upon 9 plus open parentheses y minus 1 close parentheses squared upon 4 equals 1. Here, too, a is 3 and b is 2, but the center is (3, 1). The ellipse is shown on the same graph along with the first ellipse. The center is shown to have moved 3 units right and 1 unit up.], alt: "The equation in the first figure is x squared upon 9 plus y squared upon 4 equals 1. Here, a is 3 and b is 2. The ellipse is graphed with center at (0, 0). The equation on the right is open parentheses x minus 3 close parentheses squared upon 9 plus open parentheses y minus 1 close parentheses squared upon 4 equals 1. Here, too, a is 3 and b is 2, but the center is (3, 1). The ellipse is shown on the same graph along with the first ellipse. The center is shown to have moved 3 units right and 1 unit up.", caption: none)

Notice in the graph above that we could have graphed #math.equation(block: false, alt: "the fraction open parenthesis x minus 3 close parenthesis squared over 9 plus the fraction open parenthesis y minus 1 close parenthesis squared over 4 equals 1")[$frac(attach(( x − 3 ), t: 2), 9) + frac(attach(( y − 1 ), t: 2), 4) = 1$] by translations. We moved the original ellipse to the right 3 units and then up 1 unit.

#figure(figph[This graph shows an ellipse translated from center (0, 0) to center (3, 1). The center has moved 3 units right and 1 unit up. The original ellipse has vertices at (negative 3, 0) and (3, 0) and endpoint of minor axis at (negative 2, 0) and (2, 0). The translated ellipse has vertices at (0, 1) and (6, 1) and endpoints of minor axis at (3, negative 1) and (3, 3).], alt: "This graph shows an ellipse translated from center (0, 0) to center (3, 1). The center has moved 3 units right and 1 unit up. The original ellipse has vertices at (negative 3, 0) and (3, 0) and endpoint of minor axis at (negative 2, 0) and (2, 0). The translated ellipse has vertices at (0, 1) and (6, 1) and endpoints of minor axis at (3, negative 1) and (3, 3).", caption: none)

In the next example we will use the translation method to graph the ellipse.

#examplebox("Example 5")[][
Graph #math.equation(block: false, alt: "the fraction open parenthesis x plus 4 close parenthesis squared over 16 plus the fraction open parenthesis y minus 6 close parenthesis squared over 9 equals 1")[$frac(attach(( x + 4 ), t: 2), 16) + frac(attach(( y − 6 ), t: 2), 9) = 1$] by translation.

#solutionbox[
This ellipse will have the same size and shape as #math.equation(block: false, alt: "the fraction x squared over 16 plus the fraction y squared over 9 equals 1")[$frac(x^(2), 16) + frac(y^(2), 9) = 1$] whose center is #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$] We graph this ellipse first. #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([The center is #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$]], [Center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]]),
  [Since #math.equation(block: false, alt: "16 greater than 9 ,")[$16 > 9 ,$] the major axis is horizontal.], [],
  [#math.equation(block: false, alt: "a squared equals 16 , a equals plus or minus 4")[$a^(2) = 16 , a = ± 4$] #linebreak()   #math.equation(block: false, alt: "b squared equals 9 , b equals plus or minus 3")[$b^(2) = 9 , #h(0.5em) b = ± 3$]], [The vertices are #math.equation(block: false, alt: "open parenthesis 4 , 0 close parenthesis , open parenthesis −4 , 0 close parenthesis .")[$( 4 , 0 ) , ( −4 , 0 ) .$] #linebreak() The endpoints of the minor axis are #linebreak() #math.equation(block: false, alt: "open parenthesis 0 , 3 close parenthesis , open parenthesis 0 , −3 close parenthesis .")[$( 0 , 3 ) , ( 0 , −3 ) .$]],
  [Sketch the ellipse.], [#figure(figph[A graph shows an ellipse centered at (0, 0) with x-intercepts at (-4, 0) and (4, 0), and y-intercepts at (0, 3) and (0, -3). Key points are marked.], alt: "A graph shows an ellipse centered at (0, 0) with x-intercepts at (-4, 0) and (4, 0), and y-intercepts at (0, 3) and (0, -3). Key points are marked.", caption: none)],
  [The original equation is in standard form, #linebreak() #math.equation(block: false, alt: "the fraction open parenthesis x minus h close parenthesis squared over a squared plus 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 x minus open parenthesis −4 close parenthesis close parenthesis squared over 16 plus the fraction open parenthesis y minus 6 close parenthesis squared over 9 equals 1")[$frac(attach(( x − ( −4 ) ), t: 2), 16) + frac(attach(( y − 6 ), t: 2), 9) = 1$]],
  [The ellipse is centered at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis .")[$( h , k ) .$]], [The center is #math.equation(block: false, alt: "open parenthesis −4 , 6 close parenthesis .")[$( −4 , 6 ) .$]],
  [We translate the graph of #math.equation(block: false, alt: "the fraction x squared over 16 plus the fraction y squared over 9 equals 1")[$frac(x^(2), 16) + frac(y^(2), 9) = 1$] four #linebreak() units to the left and then up 6 units. #linebreak() Verify that the center is #math.equation(block: false, alt: "open parenthesis −4 , 6 close parenthesis .")[$( −4 , 6 ) .$] #linebreak() The new ellipse is the ellipse whose equation #linebreak() is #linebreak() #math.equation(block: false, alt: "the fraction open parenthesis x plus 4 close parenthesis squared over 16 plus the fraction open parenthesis y minus 6 close parenthesis squared over 9 equals 1 .")[$frac(attach(( x + 4 ), t: 2), 16) + frac(attach(( y − 6 ), t: 2), 9) = 1 .$]], [#figure(figph[A coordinate plane displays two ellipses with labeled points. The upper ellipse has a center at (-4, 6), and the lower ellipse is centered at the origin, with dimensions indicated by arrows.], alt: "A coordinate plane displays two ellipses with labeled points. The upper ellipse has a center at (-4, 6), and the lower ellipse is centered at the origin, with dimensions indicated by arrows.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "the fraction open parenthesis x minus 5 close parenthesis squared over 9 plus the fraction open parenthesis y plus 4 close parenthesis squared over 4 equals 1")[$frac(attach(( x − 5 ), t: 2), 9) + frac(attach(( y + 4 ), t: 2), 4) = 1$] by translation.

#solutionbox[
#figure(figph[This graph shows an ellipse with center (5, negative 4), vertices (2, negative 4) and (8, negative 4) and endpoints of minor axis (5, negative 2) and (5, negative 6).], alt: "This graph shows an ellipse with center (5, negative 4), vertices (2, negative 4) and (8, negative 4) and endpoints of minor axis (5, negative 2) and (5, negative 6).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph #math.equation(block: false, alt: "the fraction open parenthesis x plus 6 close parenthesis squared over 16 plus the fraction open parenthesis y plus 2 close parenthesis squared over 25 equals 1")[$frac(attach(( x + 6 ), t: 2), 16) + frac(attach(( y + 2 ), t: 2), 25) = 1$] by translation.

#solutionbox[
#figure(figph[This graph shows an ellipse with center (negative 6, negative 2), vertices (negative 6, 3) and (negative 6, negative 7) and endpoints of minor axis (negative 10, negative 2), and (negative 2, negative 2).], alt: "This graph shows an ellipse with center (negative 6, negative 2), vertices (negative 6, 3) and (negative 6, negative 7) and endpoints of minor axis (negative 10, negative 2), and (negative 2, negative 2).", caption: none)
]
]

When an equation has both an #math.equation(block: false, alt: "x squared")[$x^(2)$] and a #math.equation(block: false, alt: "y squared")[$y^(2)$] with different coefficients, we verify that it is an ellipsis by putting it in standard form. We will then be able to graph the equation.

#examplebox("Example 6")[][
Write the equation #math.equation(block: false, alt: "x squared plus 4 y squared minus 4 x plus 24 y plus 24 equals 0")[$x^(2) + 4 y^(2) − 4 x + 24 y + 24 = 0$] in standard form and graph.

#solutionbox[
We put the equation in standard form by completing the squares in both #emph[x] and #emph[y]. #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "x squared plus 4 y squared minus 4 x plus 24 y plus 24 equals 0")[$x^(2) + 4 y^(2) − 4 x + 24 y + 24 = 0$]]),
  [Rewrite grouping the #emph[x] terms and #emph[y] terms.], [#figure(figph[A mathematical equation showing the initial steps to complete the square: (x^2 - 4x + \_) + (4y^2 + 24y + \_) = -24.], alt: "A mathematical equation showing the initial steps to complete the square: (x^2 - 4x + _) + (4y^2 + 24y + _) = -24.", caption: none)],
  [Make the coefficients of #math.equation(block: false, alt: "x squared")[$x^(2)$] and #math.equation(block: false, alt: "y squared")[$y^(2)$] equal 1.], [#figure(figph[An algebraic equation showing a step in completing the square: (x^2 - 4x + ) + 4(y^2 + 6y + ) = -24. This form is used to identify properties of conic sections.], alt: "An algebraic equation showing a step in completing the square: (x^2 - 4x + ) + 4(y^2 + 6y + ) = -24. This form is used to identify properties of conic sections.", caption: none)],
  [Complete the squares.], [#figure(figph[Mathematical equation: (x^2 - 4x + 4) + 4(y^2 + 6y + 9) = -24 + 4 + 36. This illustrates completing the square. Constants like the blue 4 and red 4, 9, 36 are highlighted, showing terms added to both sides.], alt: "Mathematical equation: (x^2 - 4x + 4) + 4(y^2 + 6y + 9) = -24 + 4 + 36. This illustrates completing the square. Constants like the blue 4 and red 4, 9, 36 are highlighted, showing terms added to both sides.", caption: none)],
  [Write as binomial squares.], [#figure(figph[The mathematical equation (x-2)^2 + 4(y+3)^2 = 16, representing an ellipse, is shown on a white background.], alt: "The mathematical equation (x-2)^2 + 4(y+3)^2 = 16, representing an ellipse, is shown on a white background.", caption: none)],
  [Divide both sides by 16 to get 1 on the right.], [#figure(figph[A mathematical equation is displayed, showing an ellipse in a non-standard form: (x-2)^2 / 16 + 4(y+3)^2 / 16 = 16 / 16. It simplifies to (x-2)^2 / 16 + (y+3)^2 / 4 = 1.], alt: "A mathematical equation is displayed, showing an ellipse in a non-standard form: (x-2)^2 / 16 + 4(y+3)^2 / 16 = 16 / 16. It simplifies to (x-2)^2 / 16 + (y+3)^2 / 4 = 1.", caption: none)],
  [Simplify.], [#figure(figph[The image displays the equation of an ellipse in standard form: (x-2)^2/16 + (y+3)^2/4 = 1.], alt: "The image displays the equation of an ellipse in standard form: (x-2)^2/16 + (y+3)^2/4 = 1.", caption: none)],
  [The equation is in standard form, #linebreak() #math.equation(block: false, alt: "the fraction open parenthesis x minus h close parenthesis squared over a squared plus the fraction open parenthesis y minus k close parenthesis squared over b squared equals 1")[$frac(attach(( x − h ), t: 2), a^(2)) + frac(attach(( y − k ), t: 2), b^(2)) = 1$]], [#figure(figph[An equation for an ellipse is displayed: the quantity x minus 2 squared over 16, plus the quantity y plus 3 squared over 4, equals 1.], alt: "An equation for an ellipse is displayed: the quantity x minus 2 squared over 16, plus the quantity y plus 3 squared over 4, equals 1.", caption: none)],
  [The ellipse is centered at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis .")[$( h , k ) .$]], [The center is #math.equation(block: false, alt: "open parenthesis 2 , −3 close parenthesis .")[$( 2 , −3 ) .$]],
  [Since #math.equation(block: false, alt: "16 greater than 4")[$16 > 4$] and 16 is in the #math.equation(block: false, alt: "x squared")[$x^(2)$] term, #linebreak() the major axis is horizontal. #linebreak()   #math.equation(block: false, alt: "a squared equals 16 , a equals plus or minus 4")[$a^(2) = 16 , a = ± 4$] #linebreak()   #math.equation(block: false, alt: "b squared equals 4 , b equals plus or minus 2")[$b^(2) = 4 , #h(0.5em) b = ± 2$]], [The distance from the center to the vertices is 4. #linebreak() The distance from the center to the endpoints of #linebreak() the minor axis is 2.],
  [Sketch the ellipse.], [#figure(figph[A graph showing an ellipse centered at (2, -3) with vertices at (-2, -3) and (6, -3), and co-vertices at (2, -1) and (2, -5).], alt: "A graph showing an ellipse centered at (2, -3) with vertices at (-2, -3) and (6, -3), and co-vertices at (2, -1) and (2, -5).", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Write the equation #math.equation(block: false, alt: "6 x squared plus 4 y squared plus 12 x minus 32 y plus 34 equals 0")[$6 x^(2) + 4 y^(2) + 12 x − 32 y + 34 = 0$] in standard form and ⓑ graph.

#solutionbox[
ⓐ #math.equation(block: false, alt: "the fraction open parenthesis x plus 1 close parenthesis squared over 6 plus the fraction open parenthesis y minus 4 close parenthesis squared over 9 equals 1")[$frac(attach(( x + 1 ), t: 2), 6) + frac(attach(( y − 4 ), t: 2), 9) = 1$] #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows an ellipse with center (negative 1, 4), vertices minus (1, 1) and (negative 1, 7) and endpoints of minor axis approximately (negative 3.5, 4) and (approximately 1.5, 4).], alt: "This graph shows an ellipse with center (negative 1, 4), vertices minus (1, 1) and (negative 1, 7) and endpoints of minor axis approximately (negative 3.5, 4) and (approximately 1.5, 4).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Write the equation #math.equation(block: false, alt: "4 x squared plus y squared minus 16 x minus 6 y plus 9 equals 0")[$4 x^(2) + y^(2) − 16 x − 6 y + 9 = 0$] in standard form and ⓑ  graph.

#solutionbox[
ⓐ #math.equation(block: false, alt: "the fraction open parenthesis x minus 2 close parenthesis squared over 4 plus the fraction open parenthesis y minus 3 close parenthesis squared over 16 equals 1")[$frac(attach(( x − 2 ), t: 2), 4) + frac(attach(( y − 3 ), t: 2), 16) = 1$] #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows an ellipse with center (2, 3), vertices (2, negative 1) and (2, 7) and endpoints of minor axis (0, 3) and (4, 3).], alt: "This graph shows an ellipse with center (2, 3), vertices (2, negative 1) and (2, 7) and endpoints of minor axis (0, 3) and (4, 3).", caption: none)
]
]


==== Solve Application with Ellipses

The orbits of the planets around the sun follow elliptical paths.

#examplebox("Example 7")[][
Pluto (a dwarf planet) moves in an elliptical orbit around the Sun. The closest Pluto gets to the Sun is approximately 30 astronomical units (AU) and the furthest is approximately 50 AU. The Sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of Pluto.

#figure(figph[This graph shows an ellipse with center (0, 0) and vertices (negative 40, 0) and (40, 0). The sun is shown at point (10, 0). This is 30 units from the right vertex and 50 units from the left vertex.], alt: "This graph shows an ellipse with center (0, 0) and vertices (negative 40, 0) and (40, 0). The sun is shown at point (10, 0). This is 30 units from the right vertex and 50 units from the left vertex.", caption: none)

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([We recognize this as an ellipse that is centered at the origin.], [#math.equation(block: false, alt: "the fraction x squared over a squared plus the fraction y squared over b squared equals 1")[$#h(2.1em) frac(x^(2), a^(2)) + frac(y^(2), b^(2)) = 1$]]),
  [Since the major axis is horizontal and the distance from the center to the vertex is 40, we know #math.equation(block: false, alt: "a equals 40")[$a = 40$] and so #math.equation(block: false, alt: "a squared equals 1600")[$a^(2) = 1600$].], [#math.equation(block: false, alt: "the fraction x squared over 1600 plus the fraction y squared over b squared equals 1")[$#h(1.1em) frac(x^(2), 1600) + frac(y^(2), b^(2)) = 1$]],
  [The minor axis is vertical but the end points aren’t given. To find #math.equation(block: false, alt: "b")[$b$] we will use the location of the Sun. Since the Sun is a focus of the ellipse at the point #math.equation(block: false, alt: "open parenthesis 10 , 0 close parenthesis")[$( 10 , 0 )$], we know #math.equation(block: false, alt: "c equals 10")[$c = 10$]. Use this to solve for #math.equation(block: false, alt: "b squared")[$b^(2)$].], [#math.equation(block: false, alt: "b squared equals a squared minus c squared; b squared equals 40 squared minus 10 squared; b squared equals 1600 minus 100; b squared equals 1500")[$#h(4.4em) b^(2) = a^(2) − c^(2) \ b^(2) = 40^(2) − 10^(2) \ b^(2) = 1600 − 100 \ b^(2) = 1500$]],
  [Substitute #math.equation(block: false, alt: "a squared")[$a^(2)$] and #math.equation(block: false, alt: "b squared")[$b^(2)$] into the standard form of the ellipse.], [#math.equation(block: false, alt: "the fraction x squared over 1600 plus the fraction y squared over 1500 equals 1")[$frac(x^(2), 1600) + frac(y^(2), 1500) = 1$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 20 AU and the furthest is approximately 30 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

#figure(figph[This graph shows an ellipse with center (0, 0) and vertices (negative 25, 0) and (25, 0). The sun is shown at point (5, 0). This is 20 units from the right vertex and 30 units from the left vertex.], alt: "This graph shows an ellipse with center (0, 0) and vertices (negative 25, 0) and (25, 0). The sun is shown at point (5, 0). This is 20 units from the right vertex and 30 units from the left vertex.", caption: none)

#solutionbox[
#math.equation(block: true, alt: "the fraction x squared over 625 plus the fraction y squared over 600 equals 1")[$frac(x^(2), 625) + frac(y^(2), 600) = 1$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 20 AU and the furthest is approximately 50 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

#figure(figph[This graph shows an ellipse with center (0, 0) and vertices (negative 35, 0) and (35, 0). The sun is shown at point (15, 0). This is 20 units from the right vertex and 50 units from the left vertex.], alt: "This graph shows an ellipse with center (0, 0) and vertices (negative 35, 0) and (35, 0). The sun is shown at point (15, 0). This is 20 units from the right vertex and 50 units from the left vertex.", caption: none)

#solutionbox[
#math.equation(block: true, alt: "the fraction x squared over 1225 plus the fraction y squared over 1000 equals 1")[$frac(x^(2), 1225) + frac(y^(2), 1000) = 1$]
]
]

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

- #link("https://openstax.org/l/37graphellipse1")[Conic Sections: Graphing Ellipses Part 1]
- #link("https://openstax.org/l/37graphellipse2")[Conic Sections: Graphing Ellipses Part 2]
- #link("https://openstax.org/l/37eqellipse")[Equation for Ellipse From Graph]
]


=== Key Concepts

- #strong[Ellipse:] An #strong[ellipse] is all points in a plane where the sum of the distances from two fixed points is constant. Each of the fixed points is called a #strong[focus] of the ellipse. #linebreak() 
#figure(figph[This figure shows two ellipses. In each, two points within the ellipse are labeled foci. A line drawn through the foci intersects the ellipse in two points. Each point is labeled a vertex. In The figure on the left, the segment connecting the vertices is called the major axis. A segment perpendicular to the major axis that passes through its midpoint and intersects the ellipse in two points is labeled minor axis. The major axis is longer than the minor axis. In The figure on the right, the segment through the foci, connecting the vertices is shorter and is labeled minor axis. Its midpoint is labeled center.], alt: "This figure shows two ellipses. In each, two points within the ellipse are labeled foci. A line drawn through the foci intersects the ellipse in two points. Each point is labeled a vertex. In The figure on the left, the segment connecting the vertices is called the major axis. A segment perpendicular to the major axis that passes through its midpoint and intersects the ellipse in two points is labeled minor axis. The major axis is longer than the minor axis. In The figure on the right, the segment through the foci, connecting the vertices is shorter and is labeled minor axis. Its midpoint is labeled center.", caption: none)

 #linebreak() 
If we draw a line through the foci intersects the ellipse in two points—each is called a #strong[vertex] of the ellipse. #linebreak() 
The segment connecting the vertices is called the #strong[major axis]. #linebreak() 
The midpoint of the segment is called the #strong[center] of the ellipse. #linebreak() 
A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the #strong[minor axis].
- #strong[Standard Form of the Equation an Ellipse with Center] #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis :")[$( 0 , 0 ) :$] The standard form of the equation of an ellipse with center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis ,")[$( 0 , 0 ) ,$] is #linebreak() 
#math.equation(block: true, alt: "the fraction x squared over a squared plus the fraction y squared over b squared equals 1")[$frac(x^(2), a^(2)) + frac(y^(2), b^(2)) = 1$] #linebreak() 
The #emph[x]-intercepts are #math.equation(block: false, alt: "open parenthesis a , 0 close parenthesis")[$( a , 0 )$] and #math.equation(block: false, alt: "open parenthesis − a , 0 close parenthesis .")[$( "−" a , 0 ) .$] #linebreak() 
The #emph[y]-intercepts are #math.equation(block: false, alt: "open parenthesis 0 , b close parenthesis")[$( 0 , b )$] and #math.equation(block: false, alt: "open parenthesis 0 , − b close parenthesis .")[$( 0 , "−" b ) .$]
- #strong[How to an Ellipse with Center] #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]
+ Write the equation in standard form.
+ Determine whether the major axis is horizontal or vertical.
+ Find the endpoints of the major axis.
+ Find the endpoints of the minor axis
+ Sketch the ellipse.
- #strong[Standard Form of the Equation an Ellipse with Center] #math.equation(block: false, alt: "open parenthesis h , k close parenthesis :")[$( h , k ) :$] The standard form of the equation of an ellipse with center #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] is #linebreak() 
#math.equation(block: true, alt: "the fraction open parenthesis x minus h close parenthesis squared over a squared plus the fraction open parenthesis y minus k close parenthesis squared over b squared equals 1")[$frac(attach(( x − h ), t: 2), a^(2)) + frac(attach(( y − k ), t: 2), b^(2)) = 1$] #linebreak() 
When #math.equation(block: false, alt: "a greater than b ,")[$a > b ,$] the major axis is horizontal so the distance from the center to the vertex is #emph[a]. #linebreak() 
When #math.equation(block: false, alt: "b greater than a ,")[$b > a ,$] the major axis is vertical so the distance from the center to the vertex is #emph[b].


==== Practice Makes Perfect

#strong[Graph an Ellipse with Center at the Origin]

In the following exercises, graph each ellipse.

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

#solutionbox[
#figure(figph[This graph shows an ellipse with center (0, 0), vertices (0, 5) and (0, negative 5) and endpoints of minor axis (2, 0) and (negative 2, 0).], alt: "This graph shows an ellipse with center (0, 0), vertices (0, 5) and (0, negative 5) and endpoints of minor axis (2, 0) and (negative 2, 0).", caption: none)
]

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

#math.equation(block: true, alt: "the fraction x squared over 25 plus the fraction y squared over 36 equals 1")[$frac(x^(2), 25) + frac(y^(2), 36) = 1$]

#solutionbox[
#figure(figph[This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (5, 0) and (negative 5, 0).], alt: "This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (5, 0) and (negative 5, 0).", caption: none)
]

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

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

#solutionbox[
#figure(figph[This graph shows an ellipse with center (0, 0), vertices (6, 0) and (negative 6, 0) and endpoints of minor axis (0, 4) and (0, negative 4).], alt: "This graph shows an ellipse with center (0, 0), vertices (6, 0) and (negative 6, 0) and endpoints of minor axis (0, 4) and (0, negative 4).", caption: none)
]

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

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

#solutionbox[
#figure(figph[This graph shows an ellipse with center (0, 0), vertices (0, 2) and (0, negative 2) and endpoints of minor axis (1, 0) and (negative 1, 0).], alt: "This graph shows an ellipse with center (0, 0), vertices (0, 2) and (0, negative 2) and endpoints of minor axis (1, 0) and (negative 1, 0).", caption: none)
]

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

#math.equation(block: true, alt: "4 x squared plus 25 y squared equals 100")[$4 x^(2) + 25 y^(2) = 100$]

#solutionbox[
#figure(figph[This graph shows an ellipse with center (0, 0), vertices (5, 0) and (negative 5, 0) and endpoints of minor axis (0, 2) and (0, negative 2).], alt: "This graph shows an ellipse with center (0, 0), vertices (5, 0) and (negative 5, 0) and endpoints of minor axis (0, 2) and (0, negative 2).", caption: none)
]

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

#math.equation(block: true, alt: "16 x squared plus 36 y squared equals 576")[$16 x^(2) + 36 y^(2) = 576$]

#solutionbox[
#figure(figph[This graph shows an ellipse with center (0, 0), vertices (6, 0) and (negative 6, 0) and endpoints of minor axis (0, 4) and (0, negative 4).], alt: "This graph shows an ellipse with center (0, 0), vertices (6, 0) and (negative 6, 0) and endpoints of minor axis (0, 4) and (0, negative 4).", caption: none)
]

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

#strong[Find the Equation of an Ellipse with Center at the Origin]

In the following exercises, find the equation of the ellipse shown in the graph.

#figure(figph[This graph shows an ellipse with center (0, 0), vertices (0, 5) and (0, negative 5) and endpoints of minor axis (negative 3, 0) and (3, 0).], alt: "This graph shows an ellipse with center (0, 0), vertices (0, 5) and (0, negative 5) and endpoints of minor axis (negative 3, 0) and (3, 0).", caption: none)

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

#figure(figph[This graph shows an ellipse with center (0, 0), vertices (5, 0) and (negative 5, 0) and endpoints of minor axis (0, 2) and (0, negative 2).], alt: "This graph shows an ellipse with center (0, 0), vertices (5, 0) and (negative 5, 0) and endpoints of minor axis (0, 2) and (0, negative 2).", caption: none)

#figure(figph[This graph shows an ellipse with center (0, 0), vertices (0, 4) and (0, negative 4) and endpoints of minor axis (negative 3, 0) and (3, 0).], alt: "This graph shows an ellipse with center (0, 0), vertices (0, 4) and (0, negative 4) and endpoints of minor axis (negative 3, 0) and (3, 0).", caption: none)

#solutionbox[
#math.equation(block: true, alt: "the fraction x squared over 9 plus the fraction y squared over 16 equals 1")[$frac(x^(2), 9) + frac(y^(2), 16) = 1$]
]

#figure(figph[This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (negative 4, 0) and (4, 0).], alt: "This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (negative 4, 0) and (4, 0).", caption: none)

#strong[Graph an Ellipse with Center Not at the Origin]

In the following exercises, graph each ellipse.

#math.equation(block: true, alt: "the fraction open parenthesis x plus 1 close parenthesis squared over 4 plus the fraction open parenthesis y plus 6 close parenthesis squared over 25 equals 1")[$frac(attach(( x + 1 ), t: 2), 4) + frac(attach(( y + 6 ), t: 2), 25) = 1$]

#solutionbox[
#figure(figph[This graph shows an ellipse with center (negative 1, negative 6, vertices (negative 1, negative 1) and (negative 1, negative 11) and endpoints of minor axis (negative 3, negative 6) and (1, negative 6).], alt: "This graph shows an ellipse with center (negative 1, negative 6, vertices (negative 1, negative 1) and (negative 1, negative 11) and endpoints of minor axis (negative 3, negative 6) and (1, negative 6).", caption: none)
]

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

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

#solutionbox[
#figure(figph[This graph shows an ellipse with center (negative 4, 2, vertices (negative 4, 5) and (negative 4, negative 1) and endpoints of minor axis (3, 1) and (negative 6, 2) and (negative 2, 2).], alt: "This graph shows an ellipse with center (negative 4, 2, vertices (negative 4, 5) and (negative 4, negative 1) and endpoints of minor axis (3, 1) and (negative 6, 2) and (negative 2, 2).", caption: none)
]

#math.equation(block: true, alt: "the fraction open parenthesis x minus 4 close parenthesis squared over 9 plus the fraction open parenthesis y minus 1 close parenthesis squared over 16 equals 1")[$frac(attach(( x − 4 ), t: 2), 9) + frac(attach(( y − 1 ), t: 2), 16) = 1$]

In the following exercises, graph each equation by translation.

#math.equation(block: true, alt: "the fraction open parenthesis x minus 3 close parenthesis squared over 4 plus the fraction open parenthesis y minus 7 close parenthesis squared over 25 equals 1")[$frac(attach(( x − 3 ), t: 2), 4) + frac(attach(( y − 7 ), t: 2), 25) = 1$]

#solutionbox[
#figure(figph[This graph shows an ellipse with center (3, 7), vertices (3, 2) and (3, 12), and endpoints of minor axis (1, 7) and (5, 7).], alt: "This graph shows an ellipse with center (3, 7), vertices (3, 2) and (3, 12), and endpoints of minor axis (1, 7) and (5, 7).", caption: none)
]

#math.equation(block: true, alt: "the fraction open parenthesis x plus 6 close parenthesis squared over 16 plus the fraction open parenthesis y plus 5 close parenthesis squared over 4 equals 1")[$frac(attach(( x + 6 ), t: 2), 16) + frac(attach(( y + 5 ), t: 2), 4) = 1$]

#math.equation(block: true, alt: "the fraction open parenthesis x minus 5 close parenthesis squared over 9 plus the fraction open parenthesis y plus 4 close parenthesis squared over 25 equals 1")[$frac(attach(( x − 5 ), t: 2), 9) + frac(attach(( y + 4 ), t: 2), 25) = 1$]

#solutionbox[
#figure(figph[This graph shows an ellipse with center (5, negative 4), vertices (5, 1) and (5, negative 9) and endpoints of minor axis (2, negative 4) and (8, negative 4).], alt: "This graph shows an ellipse with center (5, negative 4), vertices (5, 1) and (5, negative 9) and endpoints of minor axis (2, negative 4) and (8, negative 4).", caption: none)
]

#math.equation(block: true, alt: "the fraction open parenthesis x plus 5 close parenthesis squared over 36 plus the fraction open parenthesis y minus 3 close parenthesis squared over 16 equals 1")[$frac(attach(( x + 5 ), t: 2), 36) + frac(attach(( y − 3 ), t: 2), 16) = 1$]

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

#math.equation(block: true, alt: "25 x squared plus 9 y squared minus 100 x minus 54 y minus 44 equals 0")[$25 x^(2) + 9 y^(2) − 100 x − 54 y − 44 = 0$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "the fraction open parenthesis x minus 2 close parenthesis squared over 9 plus the fraction open parenthesis y minus 3 close parenthesis squared over 25 equals 1")[$frac(attach(( x − 2 ), t: 2), 9) + frac(attach(( y − 3 ), t: 2), 25) = 1$] #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows an ellipse with center (2, 3), vertices (2, negative 2) and (2, 8) and endpoints of minor axis (negative 1, 3) and (5, 3).], alt: "This graph shows an ellipse with center (2, 3), vertices (2, negative 2) and (2, 8) and endpoints of minor axis (negative 1, 3) and (5, 3).", caption: none)
]

#math.equation(block: true, alt: "4 x squared plus 25 y squared plus 8 x plus 100 y plus 4 equals 0")[$4 x^(2) + 25 y^(2) + 8 x + 100 y + 4 = 0$]

#math.equation(block: true, alt: "4 x squared plus 25 y squared minus 24 x minus 64 equals 0")[$4 x^(2) + 25 y^(2) − 24 x − 64 = 0$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "the fraction y squared over 4 plus the fraction open parenthesis x minus 3 close parenthesis squared over 25 equals 1")[$frac(y^(2), 4) + frac(attach(( x − 3 ), t: 2), 25) = 1$] #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows an ellipse with center (3, 0), vertices (negative 2, 0) and (8, 0) and endpoints of minor axis (3, 2) and (3, negative 2).], alt: "This graph shows an ellipse with center (3, 0), vertices (negative 2, 0) and (8, 0) and endpoints of minor axis (3, 2) and (3, negative 2).", caption: none)
]

#math.equation(block: true, alt: "9 x squared plus 4 y squared plus 56 y plus 160 equals 0")[$9 x^(2) + 4 y^(2) + 56 y + 160 = 0$]

In the following exercises, graph the equation.

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

#solutionbox[
#figure(figph[This graph shows a parabola with vertex (2, 1) and y intercepts (0, 0) and (2, 0).], alt: "This graph shows a parabola with vertex (2, 1) and y intercepts (0, 0) and (2, 0).", caption: none)
]

#math.equation(block: true, alt: "x squared plus y squared equals 49")[$x^(2) + y^(2) = 49$]

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

#solutionbox[
#figure(figph[This graph shows a circle with center (negative 5, negative 2) and a radius of 2 units.], alt: "This graph shows a circle with center (negative 5, negative 2) and a radius of 2 units.", caption: none)
]

#math.equation(block: true, alt: "y equals − x squared plus 8 x minus 15")[$y = "−" x^(2) + 8 x − 15$]

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

#solutionbox[
#figure(figph[This graph shows an ellipse with center (negative 3, negative 1), vertices (1, negative 1) and (negative 7, negative 1) and endpoints of minor axis (negative 3, 1) and (negative 3, negative 3).], alt: "This graph shows an ellipse with center (negative 3, negative 1), vertices (1, negative 1) and (negative 7, negative 1) and endpoints of minor axis (negative 3, 1) and (negative 3, negative 3).", caption: none)
]

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

#math.equation(block: true, alt: "the fraction x squared over 25 plus the fraction y squared over 36 equals 1")[$frac(x^(2), 25) + frac(y^(2), 36) = 1$]

#solutionbox[
#figure(figph[This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (negative 5, 0) and (5, 0).], alt: "This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (negative 5, 0) and (5, 0).", caption: none)
]

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

#math.equation(block: true, alt: "x squared plus y squared equals 64")[$x^(2) + y^(2) = 64$]

#solutionbox[
#figure(figph[This graph shows circle with center (0, 0) and with radius 8 units.], alt: "This graph shows circle with center (0, 0) and with radius 8 units.", caption: none)
]

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

#math.equation(block: true, alt: "y equals 6 x squared plus 2 x minus 1")[$y = 6 x^(2) + 2 x − 1$]

#solutionbox[
#figure(figph[This graph shows upward opening parabola. Its vertex has an x value of slightly less than 0 and a y value of slightly less than minus 1. A point on it is approximately at (negative 1, 3).], alt: "This graph shows upward opening parabola. Its vertex has an x value of slightly less than 0 and a y value of slightly less than minus 1. A point on it is approximately at (negative 1, 3).", caption: none)
]

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

#strong[Solve Application with Ellipses]

A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 10 AU and the furthest is approximately 30 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

#figure(figph[This graph shows an ellipse with center (0, 0), vertices (negative 20, 0) and (20, 0). The sun is shown at point (10, 0), which is 30 units from the left vertex and 10 units from the right vertex.], alt: "This graph shows an ellipse with center (0, 0), vertices (negative 20, 0) and (20, 0). The sun is shown at point (10, 0), which is 30 units from the left vertex and 10 units from the right vertex.", caption: none)

#solutionbox[
#math.equation(block: true, alt: "the fraction x squared over 400 plus the fraction y squared over 300 equals 1")[$frac(x^(2), 400) + frac(y^(2), 300) = 1$]
]

A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 10 AU and the furthest is approximately 70 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

#figure(figph[This graph shows an ellipse with center (0, 0), vertices (negative 40, 0) and (40, 0). The sun is shown at point (30, 0), which is 70 units from the left vertex and 10 units from the right vertex.], alt: "This graph shows an ellipse with center (0, 0), vertices (negative 40, 0) and (40, 0). The sun is shown at point (30, 0), which is 70 units from the left vertex and 10 units from the right vertex.", caption: none)

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 15 AU and the furthest is approximately 85 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

#figure(figph[This graph shows an ellipse with center (0, 0), vertices (negative 50, 0) and (50, 0). The sun is shown at point (35, 0), which is 85 units from the left vertex and 15 units from the right vertex.], alt: "This graph shows an ellipse with center (0, 0), vertices (negative 50, 0) and (50, 0). The sun is shown at point (35, 0), which is 85 units from the left vertex and 15 units from the right vertex.", caption: none)

#solutionbox[
#math.equation(block: true, alt: "the fraction x squared over 2500 plus the fraction y squared over 1275 equals 1")[$frac(x^(2), 2500) + frac(y^(2), 1275) = 1$]
]

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 15 AU and the furthest is approximately 95 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

#figure(figph[This graph shows an ellipse with center (0, 0), vertices (negative 55, 0) and (55, 0). The sun is shown at point (40, 0), which is 95 units from the left vertex and 15 units from the right vertex.], alt: "This graph shows an ellipse with center (0, 0), vertices (negative 55, 0) and (55, 0). The sun is shown at point (40, 0), which is 95 units from the left vertex and 15 units from the right vertex.", caption: none)


==== Writing Exercises

In your own words, define an ellipse and write the equation of an ellipse centered at the origin in standard form. Draw a sketch of the ellipse labeling the center, vertices and major and minor axes.

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Explain in your own words how to get the axes from the equation in standard form.

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

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Explain in your own words, the difference between a vertex and a focus of the ellipse.


==== Self Check

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

#figure(figph[This table has 4 columns 4 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first columns has the following statements: graph an ellipse with center at the origin, find the equation of an ellipse with center at the origin, graph an ellipse with center not at the origin, solve applications with ellipses. The remaining columns are blank.], alt: "This table has 4 columns 4 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first columns has the following statements: graph an ellipse with center at the origin, find the equation of an ellipse with center at the origin, graph an ellipse with center not at the origin, solve applications with ellipses. The remaining columns are blank.", caption: none)

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?