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10.1 The Parabola

Given an equation of a parabola, complete the square, if necessary, and then find the vertex, the focus, and the directrix and graph the parabola.

A conic section is formed when a right circular cone with two parts, called nappes, is intersected by a plane. One of four types of curves can be formed: a parabola, a circle, an ellipse, or a hyperbola.

Conic sections can be defined algebraically using second-degree equations of the form $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ . In addition, they can be defined geometrically as a set of points that satisfy certain conditions.

Parabolas

The graph of the quadratic function $f (x) = a x^{2} + b x + c$ $f (x) = a x^{2} + b x + c$ , $a \neq 0$ $a \neq 0$ , is a parabola. A parabola can be defined geometrically.

The line that is perpendicular to the directrix and contains the focus is the axis of symmetry. The vertex is the midpoint of the segment between the focus and the directrix. (See the figure at left.)

Let’s derive the standard equation of a parabola with vertex (0, 0) and directrix $y = - p$ $y = - p$ , where $p > 0$ $p > 0$ . We place the coordinate axes as shown in Fig. 1. The y-axis is the axis of symmetry and contains the focus F. The distance from the focus to the vertex is the same as the distance from the vertex to the directrix. Thus the coordinates of F are (0, p).

Let $P (x, y)$ $P (x, y)$ be any point on the parabola and consider $\bar{P G}$ $\bar{P G}$ perpendicular to the line $y = - p$ $y = - p$ . The coordinates of G are (x, −p). By the definition of a parabola,

\begin{array}{l} P F = P G . & The distance from P to the focus is the same \\ as the distance from P to the directrix . \end{array}

$\begin{array}{l} P F = P G . & The distance from P to the focus is the same \\ as the distance from P to the directrix . \end{array}$

Then using the distance formula, we have

\begin{array}{rcl} \sqrt{{(x - 0)}^{2} + {(y - p)}^{2}} & = & \sqrt{{(x - x)}^{2} + {[y - (- p)]}^{2}} \\ x^{2} + y^{2} - 2 p y + p^{2} & = & y^{2} + 2 p y + p^{2} & Squaring both sides and \\ squaring the binomials \\ x^{2} & = & 4 p y . \end{array}

$\begin{array}{rcl} \sqrt{{(x - 0)}^{2} + {(y - p)}^{2}} & = & \sqrt{{(x - x)}^{2} + {[y - (- p)]}^{2}} \\ x^{2} + y^{2} - 2 p y + p^{2} & = & y^{2} + 2 p y + p^{2} & Squaring both sides and \\ squaring the binomials \\ x^{2} & = & 4 p y . \end{array}$

We have shown that if $P (x, y)$ $P (x, y)$ is on the parabola shown in Fig. 1, then its coordinates satisfy this equation. The converse is also true, but we will not prove it here.

Note that if $p > 0$ $p > 0$ , as above, the graph opens up. If $p < 0$ $p < 0$ , the graph opens down.

The equation of a parabola with vertex (0, 0) and directrix $x = - p$ $x = - p$ is derived similarly. Such a parabola opens either to the right $(p > 0)$ $(p > 0)$ , as shown in Fig. 2, or to the left $(p < 0)$ $(p < 0)$ .

Example 1

Find the vertex, the focus, and the directrix of the parabola $y = - \frac{1}{12} x^{2}$ $y = - \frac{1}{12} x^{2}$ . Then graph the parabola.

Solution

We write $y = - \frac{1}{12} x^{2}$ $y = - \frac{1}{12} x^{2}$ in the form $x^{2} = 4 p y$ $x^{2} = 4 p y$ :

\begin{array}{rcl} - \frac{1}{12} x^{2} & = & y & Given equation \\ x^{2} & = & - 12 y & Multiplying by - 12 on both sides \\ x^{2} & = & 4 (- 3) y . & Standard form \end{array}

$\begin{array}{rcl} - \frac{1}{12} x^{2} & = & y & Given equation \\ x^{2} & = & - 12 y & Multiplying by - 12 on both sides \\ x^{2} & = & 4 (- 3) y . & Standard form \end{array}$

Since the equation can be written in the form $x^{2} = 4 p y$ $x^{2} = 4 p y$ , we know that the vertex is (0, 0).

We have $p = - 3$ $p = - 3$ , so the focus is (0, p), or (0, −3). The directrix is $y = - p =$ $y = - p =$ $- (- 3) = 3$ $- (- 3) = 3$ .

`x`	`y`
0	0
$\pm 1$ $\pm 1$	$- \frac{1}{12}$ $- \frac{1}{12}$
$\pm 2$ $\pm 2$	$- \frac{1}{3}$ $- \frac{1}{3}$
$\pm 3$ $\pm 3$	$- \frac{3}{4}$ $- \frac{3}{4}$
$\pm 4$ $\pm 4$	$- \frac{4}{3}$ $- \frac{4}{3}$

Now Try Exercise 7.

Example 2

Find an equation of the parabola with vertex (0, 0) and focus (5, 0). Then graph the parabola.

Solution

The focus is on the x-axis so the line of symmetry is the x-axis. Thus the equation is of the type

y^{2} = 4 p x .

$y^{2} = 4 p x .$

Since the focus (5, 0) is 5 units to the right of the vertex, $p = 5$ $p = 5$ and the equation is

y^{2} = 4 (5) x, or y^{2} = 20 x .

$y^{2} = 4 (5) x, or y^{2} = 20 x .$

`x`	$y^{2}$ $y^{2}$	`y`	(`x`, `y`)
0	0	0	(0, 0)
1	20	$\pm \sqrt{20}$ $\pm \sqrt{20}$	(1, 4.47)
			(1, −4.47)
2	40	$\pm \sqrt{40}$ $\pm \sqrt{40}$	(2, 6.32)
			(2, −6.32)
3	60	$\pm \sqrt{60}$ $\pm \sqrt{60}$	(3, 7.75)
			(3, −7.75)

Now Try Exercise 17.

Finding Standard Form by Completing the Square

If a parabola with vertex at the origin is translated horizontally $| h |$ $| h |$ units and vertically $| k |$ $| k |$ units, it has an equation as follows.

We can complete the square on equations of the form

y = a x^{2} + b x + c or x = a y^{2} + b y + c

$y = a x^{2} + b x + c or x = a y^{2} + b y + c$

in order to write them in standard form.

Example 3

For the parabola

x^{2} + 6 x + 4 y + 5 = 0,

$x^{2} + 6 x + 4 y + 5 = 0,$

find the vertex, the focus, and the directrix. Then draw the graph.

Solution

We first complete the square:

\begin{array}{rcll} x^{2} + 6 x + 4 y + 5 & = & 0 \\ x^{2} + 6 x & = & - 4 y - 5 & Subtracting 4 y and 5 on both sides \\ x^{2} + 6 x + 9 & = & - 4 y - 5 + 9 & Adding 9 on both sides to complete \\ the square on the left side \\ x^{2} + 6 x + 9 & = & - 4 y + 4 \\ {(x + 3)}^{2} & = & - 4 (y - 1) & Factoring \\ {[x - (- 3)]}^{2} & = & 4 (- 1) (y - 1) . & Writing standard form : \\ {(x - h)}^{2} = 4 p (y - k) \end{array}

$\begin{array}{rcll} x^{2} + 6 x + 4 y + 5 & = & 0 \\ x^{2} + 6 x & = & - 4 y - 5 & Subtracting 4 y and 5 on both sides \\ x^{2} + 6 x + 9 & = & - 4 y - 5 + 9 & Adding 9 on both sides to complete \\ the square on the left side \\ x^{2} + 6 x + 9 & = & - 4 y + 4 \\ {(x + 3)}^{2} & = & - 4 (y - 1) & Factoring \\ {[x - (- 3)]}^{2} & = & 4 (- 1) (y - 1) . & Writing standard form : \\ {(x - h)}^{2} = 4 p (y - k) \end{array}$

We see that $h = - 3$ $h = - 3$ , $k = 1$ $k = 1$ , and $p = - 1$ $p = - 1$ , so we have the following:

Vertex (`h`, `k`):	(−3, 1);
Focus $(h, k + p)$ $(h, k + p)$ :	$(- 3, 1 + (- 1))$ $(- 3, 1 + (- 1))$ , or (−3, 0);
Directrix $y = k - p$ $y = k - p$ :	$y = 1 - (- 1)$ $y = 1 - (- 1)$ , or $y = 2$ $y = 2$ .

Now Try Exercise 25.

Example 4

For the parabola

y^{2} - 2 y - 8 x - 31 = 0,

$y^{2} - 2 y - 8 x - 31 = 0,$

find the vertex, the focus, and the directrix. Then draw the graph.

Solution

We first complete the square:

\begin{array}{rcl} y^{2} - 2 y - 8 x - 31 & = & 0 \\ y^{2} - 2 y & = & 8 x + 31 & Adding 8 x and 31 on both sides \\ y^{2} - 2 y + 1 & = & 8 x + 31 + 1 & Adding 1 on both sides to \\ complete the square on the \\ left side \\ y^{2} - 2 y + 1 & = & 8 x + 32 \\ {(y - 1)}^{2} & = & 8 (x + 4) & Factoring \\ {(y - 1)}^{2} & = & 4 (2) [x - (- 4)] . & Writing standard form : \\ {(y - k)}^{2} = 4 p (x - h) \end{array}

$\begin{array}{rcl} y^{2} - 2 y - 8 x - 31 & = & 0 \\ y^{2} - 2 y & = & 8 x + 31 & Adding 8 x and 31 on both sides \\ y^{2} - 2 y + 1 & = & 8 x + 31 + 1 & Adding 1 on both sides to \\ complete the square on the \\ left side \\ y^{2} - 2 y + 1 & = & 8 x + 32 \\ {(y - 1)}^{2} & = & 8 (x + 4) & Factoring \\ {(y - 1)}^{2} & = & 4 (2) [x - (- 4)] . & Writing standard form : \\ {(y - k)}^{2} = 4 p (x - h) \end{array}$

We see that $h = - 4$ $h = - 4$ , $k = 1$ $k = 1$ , and $p = 2$ $p = 2$ , so we have the following:

Vertex (`h`, `k`):	(−4, 1);
Focus $(h + p, k)$ $(h + p, k)$ :	$(- 4 + 2, 1)$ $(- 4 + 2, 1)$ , or (−2, 1);
Directrix $x = h - p$ $x = h - p$ :	$x = - 4 - 2$ $x = - 4 - 2$ , or $x = - 6$ $x = - 6$ .

Now Try Exercise 31.

Applications

Parabolas have many applications. For example, cross sections of car headlights, flashlights, and searchlights are parabolas. The bulb is located at the focus and light from that point is reflected outward parallel to the axis of symmetry. Satellite dishes and field microphones used at sporting events often have parabolic cross sections. Incoming radio waves or sound waves parallel to the axis are reflected into the focus.

Similarly, in solar cooking, a parabolic mirror is mounted on a rack with a cooking pot hung in the focal area. Incoming sun rays parallel to the axis are reflected into the focus, producing a temperature high enough for cooking.

10.1 Exercise Set

In Exercises 1–6, match the equation with one of the graphs (a)–(f) that follow.

1. $x^{2} = 8 y$ $x^{2} = 8 y$
2. $y^{2} = - 10 x$ $y^{2} = - 10 x$
3. ${(y - 2)}^{2} = - 3 (x + 4)$ ${(y - 2)}^{2} = - 3 (x + 4)$
4. ${(x + 1)}^{2} = 5 (y - 2)$ ${(x + 1)}^{2} = 5 (y - 2)$
5. $13 x^{2} - 8 y - 9 = 0$ $13 x^{2} - 8 y - 9 = 0$
6. $41 x + 6 y^{2} = 12$ $41 x + 6 y^{2} = 12$

Find the vertex, the focus, and the directrix. Then draw the graph.

7. $x^{2} = 20 y$ $x^{2} = 20 y$
8. $x^{2} = 16 y$ $x^{2} = 16 y$
9. $y^{2} = - 6 x$ $y^{2} = - 6 x$
10. $y^{2} = - 2 x$ $y^{2} = - 2 x$
11. $x^{2} - 4 y = 0$ $x^{2} - 4 y = 0$
12. $y^{2} + 4 x = 0$ $y^{2} + 4 x = 0$
13. $x = 2 y^{2}$ $x = 2 y^{2}$
14. $y = \frac{1}{2} x^{2}$ $y = \frac{1}{2} x^{2}$

Find an equation of a parabola satisfying the given conditions.

15. Vertex (0, 0), focus (−3, 0)
16. Vertex (0, 0), focus (0, 10)
17. Focus (7, 0), directrix $x = - 7$ $x = - 7$
18. Focus $(0, \frac{1}{4})$ $(0, \frac{1}{4})$ , directrix $y = - \frac{1}{4}$ $y = - \frac{1}{4}$
19. Focus $(0, - π)$ $(0, - π)$ , directrix $y = π$ $y = π$
20. Focus $(- \sqrt{2}, 0)$ $(- \sqrt{2}, 0)$ , directrix $x = \sqrt{2}$ $x = \sqrt{2}$
21. Focus (3, 2), directrix $x = - 4$ $x = - 4$
22. Focus (−2, 3), directrix $y = - 3$ $y = - 3$

Find the vertex, the focus, and the directrix. Then draw the graph.

23. ${(x + 2)}^{2} = - 6 (y - 1)$ ${(x + 2)}^{2} = - 6 (y - 1)$
24. ${(y - 3)}^{2} = - 20 (x + 2)$ ${(y - 3)}^{2} = - 20 (x + 2)$
25. $x^{2} + 2 x + 2 y + 7 = 0$ $x^{2} + 2 x + 2 y + 7 = 0$
26. $y^{2} + 6 y - x + 16 = 0$ $y^{2} + 6 y - x + 16 = 0$
27. $x^{2} - y - 2 = 0$ $x^{2} - y - 2 = 0$
28. $x^{2} - 4 x - 2 y = 0$ $x^{2} - 4 x - 2 y = 0$
29. $y = x^{2} + 4 x + 3$ $y = x^{2} + 4 x + 3$
30. $y = x^{2} + 6 x + 10$ $y = x^{2} + 6 x + 10$
31. $y^{2} - y - x + 6 = 0$ $y^{2} - y - x + 6 = 0$
32. $y^{2} + y - x - 4 = 0$ $y^{2} + y - x - 4 = 0$
33. Satellite Dish. An engineer designs a satellite dish with a parabolic cross section. The dish is 15 ft wide at the opening, and the focus is placed 4 ft from the vertex.
1. Position a coordinate system with the origin at the vertex and the x-axis on the parabola’s axis of symmetry and find an equation of the parabola.
2. Find the depth of the satellite dish at the vertex.
34. Flashlight Mirror. A heavy-duty flashlight mirror has a parabolic cross section with diameter 6 in. and depth 1 in.
1. Position a coordinate system with the origin at the vertex and the x-axis on the parabola’s axis of symmetry and find an equation of the parabola.
2. How far from the vertex should the bulb be positioned if it is to be placed at the focus?
35. Spotlight. A spotlight has a parabolic cross section that is 4 ft wide at the opening and 1.5 ft deep at the vertex. How far from the vertex is the focus?
36. Ultrasound Receiver. Information Unlimited designed and sells the Ultrasonic Receiver, which detects sounds unable to be heard by the human ear. The HT90P can detect mechanical and electrical sounds such as leaking gases, air, corona, and motor friction noises. It can also be used to hear bats, insects, and even beading water. The receiver has a parabolic cross section and is 2.625 in. deep. The focus is 3.287 in. from the vertex. (Source: Information Unlimited, Amherst, NH, Robert Iannini, President) Find the diameter of the outside edge of the receiver.

Skill Maintenance

Consider the following linear equations. Without graphing them, answer the questions below.

$y = 2 x$ $y = 2 x$
$y = \frac{1}{3} x + 5$ $y = \frac{1}{3} x + 5$
$y = - 3 x - 2$ $y = - 3 x - 2$
$y = - 0.9 x + 7$ $y = - 0.9 x + 7$
$y = - 5 x + 3$ $y = - 5 x + 3$
$y = x + 4$ $y = x + 4$
$8 x - 4 y = 7$ $8 x - 4 y = 7$
$3 x + 6 y = 2$ $3 x + 6 y = 2$

37. Which has/have x-intercept $(\frac{2}{3}, 0)$ $(\frac{2}{3}, 0)$ ? [1.1]
38. Which has/have y-intercept (0, 7)? [1.1], [1.4]
39. Which slant up from left to right? [1.3]
40. Which has the least steep slant? [1.3]
41. Which has/have slope $\frac{1}{3}$ $\frac{1}{3}$ ? [1.3]
42. Which, if any, contain the point (3, 7)? [1.1]
43. Which, if any, are parallel? [1.4]
44. Which, if any, are perpendicular? [1.4]

Synthesis

45. Find an equation of the parabola with a vertical axis of symmetry and vertex (−1, 2) and containing the point (−3, 1).
46. Find an equation of a parabola with a horizontal axis of symmetry and vertex (−2, 1) and containing the point (−3, 5).
47. Suspension Bridge. The parabolic cables of a 200-ft portion of the roadbed of a suspension bridge are positioned as shown below. Vertical cables are to be spaced every 20 ft along this portion of the roadbed. Calculate the lengths of these vertical cables.

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Table of Contents for
10.1 The Parabola

10.1 The Parabola

Parabolas

Figure 1.

Figure 2.

Example 1

Solution

Example 2

Solution

Finding Standard Form by Completing the Square

Example 3

Solution

Example 4

Solution

Applications

10.1 Exercise Set

Skill Maintenance

Synthesis

Table of Contents for 10.1 The Parabola

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Table of Contents for
10.1 The Parabola