8.2 Parabolas

We begin our study of the conic sections with parabolas, since we have already seen parabolas described as graphs of quadratic functions, $f (x) = a x^{2} + b x + c$ ( $a not equal to 0$ ). It turns out that we can also describe parabolas in terms of distances.

Schematically, we have the following.

Coordinate-plane figure. — **Figure 8.16**

Each dashed line from the point $F$ to a point on the curve has the same length as the dashed line from the point on the curve to the line $D$ . The point suggestively labeled $V$ is, as you may expect, the vertex. The vertex is the point on the parabola closest to the focus.

We want to use only the distance definition of parabola to derive the equation of a parabola and, if all is right with the universe, we should get an expression much like those studied in Section 1.4.

For simplicity, assume that the vertex is $open parenthesis 0, 0 close parenthesis$ and that the parabola opens upwards. Let $p$ denote the directed¹ distance from the vertex to the focus, which by definition is the same as the distance from the vertex to the directrix. Hence, the focus is $open parenthesis 0, p close parenthesis$ and the directrix is the line $y = - p$ . Our picture becomes

From the definition of parabola, we know the distance from $open parenthesis 0, p close parenthesis$ to $(x, y)$ is the same as the distance from $(x, - p)$ to $(x, y)$ . Using the Distance Formula, Equation A.1, we get

$\begin{matrix} \sqrt{{(x - 0)}^{2} + {(y - p)}^{2}} & = & \sqrt{{(x - x)}^{2} + {(y - (- p))}^{2}} \\ \sqrt{x^{2} + {(y - p)}^{2}} & = & \sqrt{{(y + p)}^{2}} \\ x^{2} + {(y - p)}^{2} & = & {(y + p)}^{2} & square both sides \\ x^{2} + y^{2} - 2 p y + p^{2} & = & y^{2} + 2 p y + p^{2} & expand quantities \\ x^{2} & = & 4 p y & gather like terms \end{matrix}$

Solving for $y$ yields $y = \frac{x^{2}}{4 p} = \frac{1}{4 p} x^{2}$ , which is a quadratic function of the form found in Equation 1.2 with $a = \frac{1}{4 p}$ and vertex $open parenthesis 0, 0 close parenthesis$ .

We know from previous experience that if the coefficient of $x^{2}$ is negative, the parabola opens downwards. In the equation $y = \frac{1}{4 p} x^{2}$ this happens when $p less than 0$ . In our formulation, we say that $p$ is a `directed distance' from the vertex to the focus: if $p greater than 0$ , the focus is above the vertex; if $p less than 0$ , the focus is below the vertex. The focal length of a parabola, that is, the length from the vertex to the focus, is therefore $| p |$ .

If we choose to place the vertex at an arbitrary point $(h, k)$ , we arrive at the following formula using either transformations from Section 5.4 or re-deriving the formula from Definition 8.1.

Equation

The Standard Equation of a Vertical² Parabola in the $x y$ -plane:

The equation of a (vertical) parabola with vertex $(h, k)$ and focal length $| p |$ is

${(x - h)}^{2} = 4 p (y - k)$

If $p greater than 0$ , the parabola opens upwards; if $p less than 0$ , it opens downwards.

Notice that in the standard equation of the parabola above, only one of the variables, $x$ , is squared. As we'll see in the coming sections, this is a quick way to distinguish the equation of a parabola from equations representing the other conic sections.

Before embarking on an example, we take a moment to better illustrate the affect of the focal length $| p |$ on the graph of a parabola. Below we sketch three parabolas with focal length $0.5$ , $1$ , and $2$ . In each case, the focus is denoted by an ` $∗$ .' In general, as the focal length $| p |$ increases, the parabola becomes wider.

The dashed line segment in each of the illustrations above is called the latus rectum of the parabola. More specifically, the latus rectum of a parabola is the line segment with endpoints on the parabola which contains the focus and is parallel to the directrix.³

We leave it to the reader to show that the length of the latus rectum, called the focal diameter of the parabola is $vertical bar 4 ⁢ p vertical bar equals 4 ⁢ vertical bar p vertical bar$ , which appears ever so conveniently in the standard form as stated in Equation 8.1.

Knowing the focus and focal diameter allows to plot two points on the parabola in addition to the vertex, thus producing a more accurate graph.

Example 1

Graph ${(x + 1)}^{2} = - 8 (y - 3)$ in the $x y$ -plane. Find the vertex, focus, and directrix. State the focal length, focal diameter, and find the endpoints of the latus rectum.
Find the standard form of the equation of the parabola with focus $open parenthesis 2, 1 close parenthesis$ and directrix $y = - 4$ .
Find the standard form of the equation of the parabola sketched below:
Figure 8.21

Rewriting ${(x + 1)}^{2} = - 8 (y - 3)$ as ${(x - (- 1))}^{2} = - 8 (y - 3)$ , we identify $h = - 1$ and $k equals 3$ in Equation 8.1, so the vertex is $open parenthesis minus 1, 3 close parenthesis$ . Additionally, we have $4 p = - 8$ so $p = - 2$ . Since $p less than 0$ , the focus is below the vertex so the parabola opens downwards.
The focal length is $vertical bar p vertical bar equals 2$ , which means the focus is $2$ units below the vertex. From $open parenthesis minus 1, 3 close parenthesis$ , we move down $2$ units and find the focus at $(- 1, 3 - 2) = (- 1, 1)$ . Likewise the directrix is $2$ units above the vertex, or the horizontal line $y = 3 + 2 = 5$ .
The focal diameter is $| 4 p | = | - 8 | = 8$ , which means the parabola is $8$ units wide at the focus. Hence, the endpoints of the latus rectum are $4$ units to the left and right of the focus. Starting at $open parenthesis minus 1, 1 close parenthesis$ and moving to the left $4$ units, we arrive at $(- 1 - 4, 1) = (- 5, 1)$ . Starting at $open parenthesis minus 1, 1 close parenthesis$ and moving to the right $4$ units we arrive at $(- 1 + 4, 1) = (3, 1)$ . The final graph appears below on the left.
We begin by sketching the data given to us below on the right. Since the focus is $open parenthesis 2, 1 close parenthesis$ , we know the the vertex lies on the vertical line $x equals 2$ . Moreover, since the vertex is halfway between the focus and directrix, we know the vertex is exactly $\frac{5}{2}$ units below the focus at $(2, 1 - \frac{5}{2}) = (2, - \frac{3}{2})$ . This gives $h equals 2$ and $k = - \frac{3}{2}$ . Since the focus of the parabola is $\frac{5}{2}$ units above the vertex we know $p = + \frac{5}{2}$ . Using Equation 8.1, we get our final answer: ${(x - 2)}^{2} = 4 (\frac{5}{2}) {(y - (- \frac{3}{2}))}^{2}$ or ${(x - 2)}^{2} = 10 {(y + \frac{3}{2})}^{2}$ .
Figure 8.22 The graph of ${(x + 1)}^{2} = - 8 (y - 3)$
Figure 8.23
From the graph, we assume the point labeled $open parenthesis minus 1, 2 close parenthesis$ is the vertex, which means in the context of Equation 8.1, $h = - 1$ and $k equals 2$ . Hence, at this point, we know the equation is ${(x - (- 1))}^{2} = 4 p (y - 2)$ , or, more simply ${(x + 1)}^{2} = 4 p (y - 2)$ .
To determine the value of $p$ , we see $open parenthesis 1, 0 close parenthesis$ is on the graph so when $x equals 1$ , $y equals 0$ . Substituting these values into our equation gives ${(1 + 1)}^{2} = 4 p (0 - 2)$ so $4 = - 8 p$ or $p = - \frac{1}{2}$ . (The fact $p less than 0$ tracks with the parabola opening downwards.) Hence, $4 p = 4 (- \frac{1}{2}) = - 2$ so the equation of the parabola is ${(x + 1)}^{2} = - 2 (y - 2)$ . We leave it to the reader to check our answer analytically and graphically.

We can produce `horizontal' parabolas in the $x y$ -plane by reflecting our so-called `vertical' parabolas about the line $y = x$ . As you may recall from Section 5.6, we accomplish this algebraically by interchanging the variables $x$ and $y$ . Such parabolas necessarily open to the left or to the right, which means that unlike the vertical parabolas, these parabolas do not represent $y$ as a function of $x$ . As we shall see, however, they can implicitly describe $y$ as a function of $x$ , provided certain restrictions are in place.

Equation

The Standard Equation of a Horizontal Parabola:

The equation of a (horizontal) parabola with vertex $(h, k)$ and focal length $| p |$ is

${(y - k)}^{2} = 4 p (x - h)$

If $p greater than 0$ , the parabola opens to the right; if $p less than 0$ , it opens to the left.

As we saw in Section 5.6, when we reflect a horizontal line across the line $y = x$ , we obtain a vertical line, and, as a result, the directrix of a horizontal parabola is a vertical line. Moreover, the focus of a horizontal parabola is either to the left or right of the directrix. Schematically:

$p less than 0$

$p greater than 0$

Example 2

For each of the equations below:
- Graph the equation in the $x y$ -plane.
- Find the vertex, focus, and directrix. State the focal length, focal diameter, and find the endpoints of the latus rectum.
1. ${(y - 2)}^{2} = 12 (x + 1)$ .
2. $y^{2} + 4 y + 8 x = 4$
Represent each of the parabolas in number as the graphs of two or more explicit functions of $x$ .
Find the standard form of the parabola satisfying the following characteristics:
1. The focus is $open parenthesis minus 4, 2 close parenthesis$ and the directrix is the $y$ -axis.
2. The parabola whose graph is sketched below:
  Figure 8.26

1. Rewriting ${(y - 2)}^{2} = 12 (x + 1)$ as ${(y - 2)}^{2} = 12 (x - (- 1))$ , we identify $h = - 1$ and $k equals 2$ so per Equation 8.2, the vertex is $open parenthesis minus 1, 2 close parenthesis$ . We also see that $4 ⁢ p equals 12$ so $p equals 3$ . Since $p greater than 0$ , this means the focus is to the right of the vertex so the parabola opens to the right.
  The focal length is $vertical bar p vertical bar equals 3$ , which means the focus is $3$ units to the right of the vertex. From $open parenthesis minus 1, 2 close parenthesis$ , we move $3$ units to the right and find the focus at $(- 1 + 3, 2) = (2, 2)$ . Likewise the directrix is $3$ units to the left of the vertex, the vertical line $x = - 1 - 3 = - 4$ .
  The focal diameter is $| 4 p | = | 12 | = 12$ , which means the parabola is $12$ units wide at the focus. Hence, the endpoints of the latus rectum are $6$ units above and below the focus. Starting at $open parenthesis 2, 2 close parenthesis$ and moving down $6$ units, we arrive at $(2, 2 - 6) = (2, - 4)$ . Starting at $open parenthesis 2, 2 close parenthesis$ and moving up $6$ units we arrive at $(2, 2 + 6) = (2, 8)$ . The final graph appears below on the left.
2. Unlike the previous example, the equation $y^{2} + 4 y + 8 x = 4$ is not in the form described in Equation 8.2. In order to get an equivalent equation in the form prescribed by Equation 8.2, we need to complete the square in $y$ on the left-hand side of the equation. Once that is done, we factor out the coefficient of $x$ on the other side of the equation below.
  $\begin{matrix} y^{2} + 4 y + 8 x & = & 4 \\ y^{2} + 4 y & = & - 8 x + 4 \\ y^{2} + 4 y + 4 & = & - 8 x + 4 + 4 & complete the square in y . \\ {(y + 2)}^{2} & = & - 8 x + 8 & factor \\ {(y + 2)}^{2} & = & - 8 (x - 1) \end{matrix}$
  The equation ${(y + 2)}^{2} = - 8 (x - 1)$ , rewritten as ${(y - (- 2))}^{2} = - 8 (x - 1)$ is in the form given in Equation 8.2. Identifying $h equals 1$ and $k = - 2$ , we get the vertex is $open parenthesis 1, minus 2 close parenthesis$ . Moreover, we see $4 p = - 8$ so that $p = - 2$ . The fact that $p less than 0$ , means the focus will be the left of the vertex so the parabola will open to the left.
  Since the focal length is $vertical bar p vertical bar equals 2$ , the focus is $2$ units to the left of the vertex. From $open parenthesis 1, minus 2 close parenthesis$ and move left $2$ units and arrive at the focus $(1 - 2, - 2) = (- 1, - 2)$ . Similarly, the directrix is $2$ units to the right of the vertex, the vertical line $x = 1 + 2 = 3$ .
  Since the focal diameter is $vertical bar 4 ⁢ p vertical bar$ is $8$ , the parabola is $8$ units wide at the focus. Starting at the focus $(- 1, - 2)$ we move down $4$ units and get $(- 1, - 2 - 4) = (- 1, - 6)$ . Moving up $4$ units from the focus we get $(- 1, - 2 + 4) = (- 1, 2)$ . Hence, $(- 1, - 6)$ and $open parenthesis minus 1, 2 close parenthesis$ are the endpoints of the latus rectum. The final graph appears below on the right.
  Figure 8.27 The graph of ${(y - 2)}^{2} = 12 (x + 1)$
  Figure 8.28 The graph of $y^{2} + 4 y + 8 x = 4$
To describe these parabolas as graphs of functions of $x$ , we solve each equation for $y$ in terms of $x$ .
1. Starting with ${(y - 2)}^{2} = 12 (x + 1)$ , we extract square roots to get $y - 2 = \pm \sqrt{12 (x + 1)}$ . Isolating $y$ , we get $y = 2 \pm \sqrt{12 (x + 1)}$ which simplifies to $y = 2 \pm 2 \sqrt{3 x + 3}$ .
  We let $f (x) = 2 + \sqrt{3 x + 3}$ and $g (x) = 2 - \sqrt{3 x + 3}$ . Note that since $\sqrt{3 x + 3} \geq 0$ by definition, $f (x) = 2 + \sqrt{3 x + 3} \geq 2$ which means the graph of $f$ describes the upper half of the parabola. Similarly, the graph of $g (x) = 2 - \sqrt{3 x + 3}$ describes the lower half of the parabola.
2. We can solve $y^{2} + 4 y + 8 x = 4$ for $y$ by completing the square or using the quadratic formula. We leave the former to the reader,⁴ and proceed with the latter for the sake of practice.
  To use the quadratic formula, we need to set the equation to $0$ : $y^{2} + 4 y + 8 x - 4 = 0$ . Since we are solving for $y$ , we identify $a equals 1$ , $b equals 4$ and $c = 8 x - 4$ . We find the discriminant $b^{2} - 4 a c = {(4)}^{2} - 4 (1) (8 x - 4) = 32 - 32 x$ so
  $y = \frac{- 4 \pm \sqrt{32 - 32 x}}{2} = \frac{- 4 \pm \sqrt{32 (1 - x)}}{2} = \frac{- 4 \pm 4 \sqrt{2 (1 - x)}}{2} = - 2 \pm 2 \sqrt{2 - 2 x} .$
  We identify $f (x) = - 2 + 2 \sqrt{2 - 2 x}$ and $g (x) = - 2 - 2 \sqrt{2 - 2 x}$ . Since $\sqrt{2 - 2 x} \geq 0$ , we see the graph of $f$ traces out the upper half of the parabola while the graph of $g$ traces out the lower half of the parabola.
1. We sketch the data below. Since the focus is $open parenthesis minus 4, 2 close parenthesis$ and the directrix is a vertical line, we know the vertex must lie on the horizontal line $y equals 2$ . Moreover, we know the vertex must lie midway between the focus and directrix which in this case is $open parenthesis minus 2, 2 close parenthesis$ . This gives $h = - 2$ and $k equals 2$ . Since the focus is $2$ units to the left of the vertex, we know $p = - 2$ . Using Equation 8.2, we get our answer as ${(y - 2)}^{2} = 4 (- 2) (x - (- 2))$ or ${(y - 2)}^{2} = - 8 (x + 2)$ .
  Figure 8.29
2. From the graph, we may infer the vertex of the parabola is $open parenthesis minus 2, 3 close parenthesis$ so $h = - 2$ and $k equals 3$ . Per Equation 8.2, we have ${(y - 3)}^{2} = 4 p (x - (- 2))$ or ${(y - 3)}^{2} = 4 p (x + 2)$ . Since the graph contains $open parenthesis 0, 6 close parenthesis$ , we substitute $x equals 0$ and $y equals 6$ to get ${(6 - 3)}^{2} = 4 p (0 + 2)$ . We find $p = \frac{9}{8}$ which gives our final answer ${(y - 3)}^{2} = 4 (\frac{9}{8}) (x + 2)$ or ${(y - 3)}^{2} = \frac{9}{2} (x + 2)$ .

As we have seen, not all equations which describe parabolas will immediately match Equation 8.1 or Equation 8.2. Indeed, completing the square as we did with the equation in number in Example 8.2.2 will be a necessary skill not only in this section, but in the rest of this chapter.

For parabolas, we summarize the procedure for putting an equation of a parabola into standard form below. Of key importance is that in the equation for a parabola, one, and only one, of the variables are squared.

To Write the Equation of a Parabola in Standard Form

Group the variable which is squared on one side of the equation and position the non-squared variable and the constant on the other side.
Complete the square if necessary and divide by the coefficient of the perfect square.
Factor out the coefficient of the non-squared variable from it and the constant.

In studying quadratic functions, we have seen parabolas used to model physical phenomena such as the trajectories of projectiles. Other applications of the parabola concern its `reflective property' which necessitates knowing about the focus of a parabola. For example, many satellite dishes are formed in the shape of a paraboloid of revolution as depicted below.

Every cross section through the vertex of the paraboloid is a parabola with the same focus. To see why this is important, imagine the dashed lines below as electromagnetic waves heading towards a parabolic dish. It turns out that the waves reflect off the parabola and concentrate at the focus which then becomes the optimal place for the receiver.

If, on the other hand, we imagine the dashed lines as emanating from the focus, we see that the waves are reflected off the parabola in a coherent fashion as in the case in a flashlight. Here, the bulb is placed at the focus and the light rays are reflected off a parabolic mirror to give directional light.

Example 3

A satellite dish is to be constructed in the shape of a paraboloid of revolution. If the receiver placed at the focus is located 2 ft above the vertex of the dish, and the dish is to be 12 feet wide, how deep will the dish be?

One way to approach this problem is to determine the equation of the parabola suggested to us by this data. For simplicity, we'll assume the vertex is $open parenthesis 0, 0 close parenthesis$ and the parabola opens upwards. Our standard form for such a parabola is $x squared equals 4 ⁢ p ⁢ y$ . Since the focus is $2$ units above the vertex, we know $p equals 2$ , so we have $x squared equals 8 ⁢ y$ . Visually,

Figure 8.33

Since the parabola is $12$ feet wide, we know the edge is $6$ feet from the vertex. To find the depth, we are looking for the $y$ value when $x equals 6$ . Substituting $x equals 6$ into the equation of the parabola yields $6 squared equals 8 ⁢ y$ or $y = \frac{36}{8} = \frac{9}{2} = 4.5$ . Hence, the dish will be $4.5$ feet deep.

Exercises

In Exercises -, graph of the given equations in the $x y$ -plane. Find the vertex, focus and directrix. Include the endpoints of the latus rectum in your sketch.

${(x - 3)}^{2} = - 16 y$
${(x + \frac{7}{3})}^{2} = 2 (y + \frac{5}{2})$
${(y - 2)}^{2} = - 12 (x + 3)$
${(y + 4)}^{2} = 4 x$
${(x - 1)}^{2} = 4 (y + 3)$
${(x + 2)}^{2} = - 20 (y - 5)$
${(y - 4)}^{2} = 18 (x - 2)$
${(y + \frac{3}{2})}^{2} = - 7 (x + \frac{9}{2})$

In Exercises -, put the equation into standard form. Find the vertex, focus and directrix.⁵

$y^{2} - 10 y - 27 x + 133 = 0$
$25 x^{2} + 20 x + 5 y - 1 = 0$
$x^{2} + 2 x - 8 y + 49 = 0$
$2 y^{2} + 4 y + x - 8 = 0$
$x^{2} - 10 x + 12 y + 1 = 0$
$3 y^{2} - 27 y + 4 x + \frac{211}{4} = 0$
For each of the equations given in Exercises - that do not describe $y$ as a function of $x$ , find two or more explicit functions of $x$ represented by each of the equations. (See Example 8.2.2.)

In Exercises -, find an equation for the parabola whose graph is given.

Figure 8.34
Figure 8.35
Figure 8.36 $x$ , $y$ -intercept $open parenthesis 0, 0 close parenthesis$
Figure 8.37 $x$ -intercept $open parenthesis 1, 0 close parenthesis$ , $y$ -intercept $open parenthesis 0, 2 close parenthesis$

In Exercises -, find an equation for the parabola which fits the given criteria.

Vertex $open parenthesis 7, 0 close parenthesis$ , focus $open parenthesis 0, 0 close parenthesis$ .
Focus $open parenthesis 10, 1 close parenthesis$ , directrix $x equals 5$ .
Vertex $(- 8, - 9)$ ; $open parenthesis 0, 0 close parenthesis$ and $open parenthesis minus 16, 0 close parenthesis$ are points on the curve.
The endpoints of latus rectum are $(- 2, - 7)$ and $open parenthesis 4, minus 7 close parenthesis$ .
The mirror in Carl's flashlight is a paraboloid of revolution. If the mirror is 5 centimeters in diameter and 2.5 centimeters deep, where should the light bulb be placed so it is at the focus of the mirror?
A parabolic Wi-Fi antenna is constructed by taking a flat sheet of metal and bending it into a parabolic shape.⁶ If the cross section of the antenna is a parabola which is 45 centimeters wide and 25 centimeters deep, where should the receiver be placed to maximize reception?
A parabolic arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch exactly 1 foot in from the base of the arch.
A popular novelty item is the `mirage bowl.' Follow this link to see another startling application of the reflective property of the parabola.
With the help of your classmates, research spinning liquid mirrors. To get you started, here .

Answers

${(x - 3)}^{2} = - 16 y$ Vertex $open parenthesis 3, 0 close parenthesis$ Focus $open parenthesis 3, minus 4 close parenthesis$ Directrix $y equals 4$ Endpoints of latus rectum $(- 5, - 4)$ , $open parenthesis 11, minus 4 close parenthesis$
Figure 8.38
${(x + \frac{7}{3})}^{2} = 2 (y + \frac{5}{2})$ Vertex $(- \frac{7}{3}, - \frac{5}{2})$ Focus $(- \frac{7}{3}, - 2)$ Directrix $y = - 3$ Endpoints of latus rectum $(- \frac{10}{3}, - 2)$ , $(- \frac{4}{3}, - 2)$
Figure 8.39
${(y - 2)}^{2} = - 12 (x + 3)$ Vertex $open parenthesis minus 3, 2 close parenthesis$ Focus $open parenthesis minus 6, 2 close parenthesis$ Directrix $x equals 0$ Endpoints of latus rectum $open parenthesis minus 6, 8 close parenthesis$ , $(- 6, - 4)$
Figure 8.40
${(y + 4)}^{2} = 4 x$ Vertex $open parenthesis 0, minus 4 close parenthesis$ Focus $open parenthesis 1, minus 4 close parenthesis$ Directrix $x = - 1$ Endpoints of latus rectum $open parenthesis 1, minus 2 close parenthesis$ , $open parenthesis 1, minus 6 close parenthesis$
Figure 8.41
${(x - 1)}^{2} = 4 (y + 3)$ Vertex $open parenthesis 1, minus 3 close parenthesis$ Focus $open parenthesis 1, minus 2 close parenthesis$ Directrix $y = - 4$ Endpoints of latus rectum $open parenthesis 3, minus 2 close parenthesis$ , $(- 1, - 2)$
Figure 8.42
${(x + 2)}^{2} = - 20 (y - 5)$ Vertex $open parenthesis minus 2, 5 close parenthesis$ Focus $open parenthesis minus 2, 0 close parenthesis$ Directrix $y equals 10$ Endpoints of latus rectum $open parenthesis minus 12, 0 close parenthesis$ , $open parenthesis 8, 0 close parenthesis$
Figure 8.43
${(y - 4)}^{2} = 18 (x - 2)$ Vertex $open parenthesis 2, 4 close parenthesis$ Focus $(\frac{13}{2}, 4)$ Directrix $x = - \frac{5}{2}$ Endpoints of latus rectum $(\frac{13}{2}, - 5)$ , $(\frac{13}{2}, 13)$
Figure 8.44
${(y + \frac{3}{2})}^{2} = - 7 (x + \frac{9}{2})$ Vertex $(- \frac{9}{2}, - \frac{3}{2})$ Focus $(- \frac{25}{4}, - \frac{3}{2})$ Directrix $x = - \frac{11}{4}$ Endpoints of latus rectum $(- \frac{25}{4}, 2)$ , $(- \frac{25}{4}, - 5)$
Figure 8.45
${(y - 5)}^{2} = 27 (x - 4)$ Vertex $open parenthesis 4, 5 close parenthesis$ Focus $(\frac{43}{4}, 5)$ Directrix $x = - \frac{11}{4}$
${(x + \frac{2}{5})}^{2} = - \frac{1}{5} (y - 1)$ Vertex $(- \frac{2}{5}, 1)$ Focus $(- \frac{2}{5}, \frac{19}{20})$ Directrix $y = \frac{21}{20}$
${(x + 1)}^{2} = 8 (y - 6)$ Vertex $open parenthesis minus 1, 6 close parenthesis$ Focus $open parenthesis minus 1, 8 close parenthesis$ Directrix $y equals 4$
${(y + 1)}^{2} = - \frac{1}{2} (x - 10)$ Vertex $open parenthesis 10, minus 1 close parenthesis$ Focus $(\frac{79}{8}, - 1)$ Directrix $x = \frac{81}{8}$
${(x - 5)}^{2} = - 12 (y - 2)$ Vertex $open parenthesis 5, 2 close parenthesis$ Focus $open parenthesis 5, minus 1 close parenthesis$ Directrix $y equals 5$
${(y - \frac{9}{2})}^{2} = - \frac{4}{3} (x - 2)$ Vertex $(2, \frac{9}{2})$ Focus $(\frac{5}{3}, \frac{9}{2})$ Directrix $x = \frac{7}{3}$
The equations which do not represent $y$ as a function of $x$ are:,,,,,,.
For number:
- $f (x) = 2 + 2 \sqrt{- 3 x - 9}$ represents the upper half of the parabola.
- $g (x) = 2 - 2 \sqrt{- 3 x - 9}$ represents the lower half of the parabola.
For number:
- $f (x) = - 4 + 2 \sqrt{x}$ represents the upper half of the parabola.
- $g (x) = - 4 - 2 \sqrt{x}$ represents the lower half of the parabola.
For number:
- $f (x) = 4 + 3 \sqrt{2 x - 4}$ represents the upper half of the parabola.
- $g (x) = 4 - 3 \sqrt{2 x - 4}$ represents the lower half of the parabola.
For number:
- $f (x) = - \frac{3}{2} + \frac{1}{2} \sqrt{- 28 x - 126}$ represents the upper half of the parabola.
- $g (x) = - \frac{3}{2} - \frac{1}{2} \sqrt{- 28 x - 126}$ represents the lower half of the parabola.
For number:
- $f (x) = 5 + 3 \sqrt{3 x - 12}$ represents the upper half of the parabola.
- $g (x) = 5 - 3 \sqrt{3 x - 12}$ represents the lower half of the parabola.
For number:
- $f (x) = - 1 + \frac{1}{2} \sqrt{- 2 x + 20}$ represents the upper half of the parabola.
- $g (x) = - 1 - \frac{1}{2} \sqrt{- 2 x + 20}$ represents the lower half of the parabola.
For number:
- $f (x) = \frac{9}{2} + \frac{2}{3} \sqrt{- 3 x + 6}$ represents the upper half of the parabola.
- $f (x) = \frac{9}{2} - \frac{2}{3} \sqrt{- 3 x + 6}$ represents the lower half of the parabola.
${(x + 2)}^{2} = \frac{1}{2} (y + 6)$
$x^{2} = - \frac{1}{2} (y - 4)$
${(y - 2)}^{2} = x + 4$
$y^{2} = - 16 (x - 1)$
$y^{2} = - 28 (x - 7)$
${(y - 1)}^{2} = 10 (x - \frac{15}{2})$
${(x + 8)}^{2} = \frac{64}{9} (y + 9)$
${(x - 1)}^{2} = 6 (y + \frac{17}{2})$ or ${(x - 1)}^{2} = - 6 (y + \frac{11}{2})$
The bulb should be placed $0.625$ centimeters above the vertex of the mirror.⁷
The receiver should be placed $5.0625$ centimeters from the vertex of the cross section of the antenna.
The arch can be modeled by $x^{2} = - (y - 9)$ or $y = 9 - x^{2}$ . One foot in from the base of the arch corresponds to either $x = \pm 2$ , so the height is $y = 9 - {(\pm 2)}^{2} = 5$ feet.

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.