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Section 7.3 The Ellipse

Before Starting this Section, Review

1 Distance formula (Section 2.1 , page 178)
2 Completing the square (Section 3.1 , page 321)
3 Midpoint formula (Section 2.1 , page 180)
4 Transformations of graphs (Section 2.7 , page 274)
5 Symmetry (Section 2.2 , page 188)

Objectives

1 Define an ellipse.
2 Find the equation of an ellipse.
3 Translate ellipses.
4 Use ellipses in applications.

What Is Lithotripsy?

Lithotripsy is a combination of two words: litho and tripsy. In Greek, the word lith denotes a stone and the word tripsy means “crushing.” The complete medical term for lithotripsy is extracorporeal shock-wave lithotripsy (ESWL). Lithotripsy is a medical procedure in which a kidney stone is crushed into small sandlike pieces with the help of ultrasound high-energy shock waves, without breaking the skin.

The lithotripsy machine is called a lithotripter. The technology for the lithotripter was developed in Germany. The first successful treatment of a patient by a lithotripter was in February 1980 at Munich University. In the beginning, the patient had to lie in an elliptical tank filled with water and only small kidney stones were crushed. But as medical research advanced, newer machines with better technology were manufactured. Now there is no need to keep the patient unconscious with an empty stomach, nor is the tub of water required.

In Example 6, you will see how the reflecting property of an ellipse is used in lithotripsy.

Definition of Ellipse

1 Define an ellipse.

You can draw an ellipse with the use of thumbtacks, a fixed length of string, and a pencil. Stick a thumbtack at each of the two points (the foci) and tie one end of the string to each tack. Place a pencil inside the loop of the string and pull it taut. Move the pencil, keeping the string taut at all times. The pencil traces an ellipse, as shown in Figure 7.17.

The points of intersection of the ellipse with the line through the foci are called the vertices (plural of vertex). The line segment connecting the vertices is the major axis of the ellipse. The midpoint of the major axis is the center of the ellipse. The line segment that is perpendicular to the major axis at the center and with endpoints on the ellipse is called the minor axis. See Figure 7.18.

Equation of an Ellipse

2 Find the equation of an ellipse.

To find a simple equation of an ellipse, we place the ellipse’s major axis along one of the coordinate axes and the center of the ellipse at the origin. We will begin with the case in which the major axis lies along the x-axis. Let F1(−c, 0) $F_{1} (- c, 0)$ and F2(c, 0) $F_{2} (c, 0)$ be the coordinates of the foci. See Figure 7.19. Note that the center at the origin is the midpoint of the line segment having the foci as endpoints. To simplify the equation, it is customary to let 2a be the constant distance referred to in the definition. Let P(x, y) be a point on the ellipse. Then

d (P, F 1) + d (P, F 2) (x + c) 2 + y 2 - - - - - - - - - - - \sqrt + (x - c) 2 + y 2 - - - - - - - - - - - \sqrt (x + c) 2 + y 2 - - - - - - - - - - - \sqrt (x + c) 2 + y 2 x 2 + 2 c x + c 2 + y 2 x 2 + 2 c x + c 2 + y 2 4 c x - 4 a 2 c x - a 2 (c x - a 2) 2 c 2 x 2 - 2 a 2 c x + a 4 (c 2 - a 2) x 2 - a 2 y 2 (a 2 - c 2) x 2 + a 2 y 2 = = = = = = = = = = = = 2 a 2 a 2 a - (x - c) 2 + y 2 - - - - - - - - - - - \sqrt (2 a - (x - c) 2 + y 2 - - - - - - - - - - - \sqrt) 2 4 a 2 - 4 a (x - c) 2 + y 2 - - - - - - - - - - - \sqrt + (x - c) 2 + y 2 4 a 2 - 4 a (x - c) 2 + y 2 - - - - - - - - - - - \sqrt + x 2 - 2 c x + c 2 + y 2 - 4 a (x - c) 2 + y 2 - - - - - - - - - - - \sqrt - a (x - c) 2 + y 2 - - - - - - - - - - - \sqrt a 2 [(x - c) 2 + y 2] a 2 (x 2 - 2 c x + c 2 + y 2) a 2 c 2 - a 4 a 2 (a 2 - c 2) (1) Definition of ellipse Distance formula Isolate a radical . Square both sides . Expand squares . Expand squares . Isolate the radical and simplify . Divide both sides by 4. Square both sides . Expand squares . Simplify and rearrange . Multiply both sides by - 1 and factor a 4 - a 2 c 2 .

$\begin{array}{rcll} d (P, F_{1}) + d (P, F_{2}) & = & 2 a & Definition of ellipse \\ \sqrt{{(x + c)}^{2} + y^{2}} + \sqrt{{(x - c)}^{2} + y^{2}} & = & 2 a & Distance formula \\ \sqrt{{(x + c)}^{2} + y^{2}} & = & 2 a - \sqrt{{(x - c)}^{2} + y^{2}} & Isolate a radical . \\ {(x + c)}^{2} + y^{2} & = & {(2 a - \sqrt{{(x - c)}^{2} + y^{2}})}^{2} & Square both sides . \\ x^{2} + 2 c x + c^{2} + y^{2} & = & 4 a^{2} - 4 a \sqrt{{(x - c)}^{2} + y^{2}} \\ + {(x - c)}^{2} + y^{2} & Expand squares . \\ x^{2} + 2 c x + c^{2} + y^{2} & = & 4 a^{2} - 4 a \sqrt{{(x - c)}^{2} + y^{2}} \\ + x^{2} - 2 c x + c^{2} + y^{2} & Expand squares . \\ 4 c x - 4 a^{2} & = & - 4 a \sqrt{{(x - c)}^{2} + y^{2}} & Isolate the radical and simplify . \\ c x - a^{2} & = & - a \sqrt{{(x - c)}^{2} + y^{2}} & Divide both sides by 4. \\ {(c x - a^{2})}^{2} & = & a^{2} [{(x - c)}^{2} + y^{2}] & Square both sides . \\ c^{2} x^{2} - 2 a^{2} c x + a^{4} & = & a^{2} (x^{2} - 2 c x + c^{2} + y^{2}) & Expand squares . \\ (c^{2} - a^{2}) x^{2} - a^{2} y^{2} & = & a^{2} c^{2} - a^{4} & Simplify and rearrange . \\ (a^{2} - c^{2}) x^{2} + a^{2} y^{2} & = & a^{2} (a^{2} - c^{2}) (1) & Multiply both sides by - 1 and factor a^{4} - a^{2} c^{2} . \end{array}$

Now consider triangle F1PF2 $F_{1} P F_{2}$ in Figure 7.19. The Triangle Inequality from geometry states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side; so for this triangle, we have

d (P, F 1) + d (P, F 2) 2 a a a 2 > > > > d (F 1, F 2) 2 c c c 2 2 a = d (P, F 1) + d (P, F 2); d (F 1, F 2) = 2 c Divide both sides by 2. Square both sides .

$\begin{array}{rcll} d (P, F_{1}) + d (P, F_{2}) & > & d (F_{1}, F_{2}) \\ 2 a & > & 2 c & 2 a = d (P, F_{1}) + d (P, F_{2}); d (F_{1}, F_{2}) = 2 c \\ a & > & c & Divide both sides by 2. \\ a^{2} & > & c^{2} & Square both sides . \end{array}$

Figure 7.19

Deriving the equation of an ellipse

Therefore, a2−c2>0. $a^{2} - c^{2} > 0 .$ Letting b2=a2−c2, $b^{2} = a^{2} - c^{2},$ we obtain

(a 2 - c 2) x 2 + a 2 y 2 b 2 x 2 + a 2 y 2 x 2 a 2 + y 2 b 2 = = = a 2 (a 2 - c 2) a 2 b 2 1 (2) Equation (1) Replace a 2 - c 2 with b 2 . Divide both sides by a 2 b 2 .

$\begin{array}{rcll} (a^{2} - c^{2}) x^{2} + a^{2} y^{2} & = & a^{2} (a^{2} - c^{2}) & Equation (1) \\ b^{2} x^{2} + a^{2} y^{2} & = & a^{2} b^{2} & Replace a^{2} - c^{2} with b^{2} . \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} & = & 1 (2) & Divide both sides by a^{2} b^{2} . \end{array}$

The last equation is the standard form of the equation of an ellipse with center (0, 0) and foci (−c, 0) $(- c, 0)$ and (c, 0), where b2=a2−c2. $b^{2} = a^{2} - c^{2} .$

To find the x-intercepts for the graph of this ellipse, set y=0 $y = 0$ in equation (2) to get x2a2=1. $\frac{x^{2}}{a^{2}} = 1 .$

The solutions are x=±a, $x = \pm a,$ so the x-intercepts of the graph are −a $- a$ and a.

Therefore, the ellipse’s vertices are (−a, 0) $(- a, 0)$ and (a, 0). The distance between the vertices V1(−a, 0) $V_{1} (- a, 0)$ and V2(a, 0) $V_{2} (a, 0)$ is 2a, so the length of the major axis is 2a.

The y-intercepts of the ellipse (found by setting x=0 $x = 0$ in equation (2)) are −b $- b$ and b, so the endpoints of the minor axis are (0, −b) $(0, - b)$ and (0, b) Consequently, the length of the minor axis is 2b. See the left-hand figure in the table below.

Similarly, by reversing the roles of x and y, an equation of the ellipse with center (0, 0) and foci (0, −c) $(0, - c)$ and (0, c) on the y-axis is given by

x 2 b 2 + y 2 a 2 = 1, where b 2 = a 2 - c 2 (3)

$\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1, where b^{2} = a^{2} - c^{2} (3)$

In equation (3), the major axis, of length 2a, is along the y-axis and the minor axis, of length 2b, is along the x-axis. See the right-hand figure in the table below.

If the major axis of an ellipse is along or parallel to the x-axis, the ellipse is called a horizontal ellipse, an ellipse with major axis along or parallel to the y-axis is called a vertical ellipse.

We summarize the facts about horizontal and vertical ellipses with center (0, 0). By the equation of a major or minor axis, we mean the equation of the line on which that axis lies.

Summary of Main Facts

Main facts about an ellipse with center (0, 0)
Standard Equation	x2a2+y2b2=1; a>b>0 $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1; a > b > 0$ (Horizontal ellipse)	$\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1; a > b > 0$ (Vertical ellipse)
Relationship between `a`, `b`, and `c`	$b^{2} = a^{2} - c^{2}$	$b^{2} = a^{2} - c^{2}$
Major axis along	`x`-axis	`y`-axis
Length of major axis	2`a`	2`a`
Minor axis along	`y`-axis	`x`-axis
Length of minor axis	2`b`	2`b`
Vertices	$(\pm a, 0)$	$(0, \pm a)$
Foci	$(\pm c, 0)$	$(0, \pm c)$
Endpoints of minor axis	$(0, \pm b)$	$(\pm b, 0)$
Symmetry	The graph is symmetric about the `x`-axis, `y`-axis, and origin.	The graph is symmetric about the `x`-axis, `y`-axis, and origin.
Graph

Notice that if $a = b,$ then $a^{2} = b^{2}$ and equations (2) and (3) of an ellipse are equivalent to $x^{2} + y^{2} = a^{2};$ this is the equation of a circle with center at (0, 0) and radius a. Consequently, a circle is a special case of an ellipse.

Example 1 Finding an Equation of an Ellipse

Find the standard form of the equation of the ellipse that has vertex (5, 0) and foci $(\pm 4, 0) .$

Solution

Because the foci are $(- 4, 0)$ and (4, 0), the major axis is on the x-axis. We know that $c = 4$ and $a = 5$ . Then find $b^{2} :$

$\begin{array}{l} b^{2} = a^{2} - c^{2} & Relationship between a, b, and c \\ b^{2} = {(5)}^{2} - {(4)}^{2} = 9 & Replace c with 4 and a with 5. \end{array}$

Substituting 25 for $a^{2}$ and 9 for $b^{2}$ in the standard form for a horizontal ellipse, we get

$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 .$

Practice Problem 1

Find the standard form of the equation of the ellipse that has vertex (0, 10) and foci $(0, - 8)$ and (0, 8).

Example 2 Graphing an Ellipse

Graph the ellipse whose equation is $9 x^{2} + 4 y^{2} = 36 .$ Find the foci of the ellipse and the lengths of its axes.

Solution

First, write the equation in standard form.

Because the denominator in the $y^{2}$ -term is larger than the denominator in the $x^{2}$ -term, the ellipse is a vertical ellipse. Here $a^{2} = 9$ and $b^{2} = 4,$ so $c^{2} = a^{2} - b^{2} = 9 - 4 = 5 .$ Thus, $a = 3, b = 2$ and $c = \sqrt{5} .$ From the table on page 678, we know the following features of the vertical ellipse:

$\begin{array}{lrcl} Vertices: & (0, \pm a) & = & (0, \pm 3) \\ Foci: & (0, \pm c) & = & (0, \pm \sqrt{5}) \\ Length of major axis: & 2 a & = & 2 (3) = 6 \\ Length of minor axis: & 2 b & = & 2 (2) = 4 \end{array}$

The graph of the ellipse is shown in Figure 7.20.

Figure 7.20

Ellipse with equation $9 x^{2} + 4 y^{2} = 36$

Practice Problem 2

Repeat Example 2 for the ellipse whose equation is $4 x^{2} + y^{2} = 16 .$

Translations of Ellipses

3 Translate ellipses

The graph of the equation

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

is an ellipse with center (0, 0) and major axis along a coordinate axis. As in the case of a parabola, horizontal and vertical shifts can be used to obtain the graph of an ellipse whose equation is

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1 .$

The center of such an ellipse is (h, k), and its axes are parallel to the coordinate axes.

Summary of Main Facts

Main facts about horizontal and vertical ellipses with center (`h`, `k`)
Standard Equation	$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1; a > b > 0$ (Horizontal ellipse)	$\frac{{(x - h)}^{2}}{b^{2}} + \frac{{(y - k)}^{2}}{a^{2}} = 1; a > b > 0$ (Vertical ellipse)
Center	(`h`, `k`)	(`h`, `k`)
Major axis along the line	$y = k$	$x = h$
Length of major axis	2`a`	2`a`
Minor axis along the line	$x = h$	$y = k$
Length of minor axis	2`b`	2`b`
Vertices	$(h + a, k), (h - a, k)$	$(h, k + a), (h, k - a)$
Endpoints of minor axis	$(h, k - b), (h, k + b)$	$(h - b, k), (h + b, k)$ ,
Foci	$(h + c, k), (h - c, k)$	$(h, k + c), (h, k - c)$
Relations between `a`, `b`, and `c`	$c^{2} = a^{2} - b^{2}$	$c^{2} = a^{2} - b^{2}$
Symmetry	The graph is symmetric about the lines $x = h$ and $y = k .$	The graph is symmetric about the lines $x = h$ and $y = k .$
Graph

Example 3 Finding the Equation of an Ellipse

Find an equation of the ellipse that has

foci $(- 3, 2)$ and (5, 2), and has a major axis of length 10.
Vertices $(- 1, 2)$ and $(- 1, - 2)$ and a focus at $(- 1, - 1)$ .

Solution

Since the foci $(- 3, 2)$ and (5, 2) lie on the horizontal line $y = 2,$ the ellipse is a horizontal ellipse. The center of the ellipse is the midpoint of the line segment joining the foci. Using the midpoint formula yields

$h = \frac{- 3 + 5}{2} = 1 k = \frac{2 + 2}{2} = 2 .$

The center of the ellipse is (1, 2). See Figure 7.21. Since the length of the major axis is 10, the vertices must be at a distance $a = 5$ units from the center.

Figure 7.21

Finding the center given the foci

In Figure 7.21, the foci are 4 units from the center. Thus, $c = 4 .$ Now find $b^{2} :$

$b^{2} = a^{2} - c^{2} = {(5)}^{2} - {(4)}^{2} = 9$

Since the major axis is horizontal, the standard form of the equation of the ellipse is

$\begin{array}{l} \frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1 & a > b > 0 \\ \frac{{(x - 1)}^{2}}{25} + \frac{{(y - 2)}^{2}}{9} = 1 & \begin{array}{l} Replace h with 1, k with 2, \\ a^{2} with 25, and b^{2} with 9. \end{array} \end{array}$

Now for this horizontal ellipse with center (1, 2), we have $a = 5, b = 3,$ and $c = 4 .$

$\begin{array}{c} Vertices \\ (h \pm a, k) = (1 \pm 5, 2) = (- 4, 2) and (6, 2) \\ Endpoints of Minor Axis \\ (h, k \pm b) = (1, 2 \pm 3) = (1, - 1) and (1, 5) \end{array}$

Use the center and these four points to sketch the graph shown in Figure 7.22.

Figure 7.22

Ellipse with foci $(- 3, 2)$ and (5, 2)
Since the vertices $(- 1, 2)$ and $(- 1, - 2)$ lie on the vertical line $x = - 1,$ the ellipse is a vertical ellipse. The center (h, k) of the ellipse is the midpoint of the line segment joining the vertices. As in part a, use the midpoint formula to find the center (h, k), with $h = - 1$ and $k = 0$ . The distance, a, from the center $(- 1, 0)$ to the vertex $(- 1, 2)$ is $\sqrt{{(- 1 - (- 1))}^{2} + {(0 - 2)}^{2}} = \sqrt{0^{2} + {(- 2)}^{2}} = \sqrt{4} = 2,$ so $a = 2$ . The distance, c, from the center $(- 1, 0)$ to the focus $(- 1, - 1)$ is $\sqrt{{(- 1 - (- 1))}^{2} + {(0 - (- 1))}^{2}} = \sqrt{0^{2} + 1^{2}} = \sqrt{1} = 1,$ so $c = 1$ . Then,

$\begin{array}{l} b^{2} = a^{2} - c^{2} & Relationship involving a, b, and c . \\ b^{2} = {(2)}^{2} - 1^{2} = 4 - 1 = 3 & Replace c with 1 and a with 2. \\ \frac{{(x - h)}^{2}}{b^{2}} + \frac{{(y - k)}^{2}}{a^{2}} = 1 & Equation for a vertical ellipse . \\ \frac{{(x - [- 1])}^{2}}{3} + \frac{{(y - 0)}^{2}}{4} = 1 & \begin{array}{l} Replace h with - 1, b^{2} with 3, \\ k with 0, and a^{2} with 4. \end{array} \\ \frac{{(x + 1)}^{2}}{3} + \frac{y^{2}}{4} = 1 & Simplify . \end{array}$

Practice Problem 3

Find an equation of the ellipse that has
1. foci at $(2, - 3)$ and (2, 5) and has a major axis of length 10
2. vertices at (0, 0) and (0, 10) and a focus at (0, 8).

Both standard forms of the equations of an ellipse with center (h, k) can be put into the form

$A x^{2} + C y^{2} + D x + E y + F = 0 (4)$

where neither A nor C is zero and both have the same sign. For example, consider the ellipse with equation

$\begin{array}{rcll} \frac{{(x - 1)}^{2}}{4} + \frac{{(y + 2)}^{2}}{9} & = & 1. \\ 9 {(x - 1)}^{2} + 4 {(y + 2)}^{2} & = & 36 & Multiply both sides by 36. \\ 9 (x^{2} - 2 x + 1) + 4 (y^{2} + 4 y + 4) & = & 36 & Expand squares . \\ 9 x^{2} - 18 x + 9 + 4 y^{2} + 16 y + 16 & = & 36 & Distributive property \\ 9 x^{2} + 4 y^{2} - 18 x + 16 y - 11 & = & 0 & Simplify . \end{array}$

Comparing this equation with equation (4), we have

$A = 9, C = 4, D = - 18, E = 16, and F = - 11.$

Conversely, equation (4) can be put into a standard form for an ellipse by completing the squares in both of the variables x and y.

Example 4 Converting to Standard Form

Find the center, vertices, and foci of the ellipse with equation

$3 x^{2} + 4 y^{2} + 12 x - 8 y - 32 = 0 .$

Solution

Complete squares on x and y.

$\begin{array}{rcll} 3 x^{2} + 4 y^{2} + 12 x - 8 y - 32 & = & 0 & Original equation \\ (3 x^{2} + 12 x) + (4 y^{2} - 8 y) & = & 32 & Group like terms . \\ 3 (x^{2} + 4 x) + 4 (y^{2} - 2 y) & = & 32 & Factor out 3 and 4. \\ 3 (x^{2} + 4 x + 4) + 4 (y^{2} - 2 y + 1) & = & 32 + 12 + 4 & Complete squares on x and y : \\ add 3 \cdot 4 and 4 \cdot 1. \\ 3 {(x + 2)}^{2} + 4 {(y - 1)}^{2} & = & 48 & Simplify . \\ \frac{{(x + 2)}^{2}}{16} + \frac{{(y - 1)}^{2}}{12} & = & 1 & \begin{array}{l} Divide both sides by 48. (16 is \\ the larger denominator .) \end{array} \end{array}$

The last equation is the standard form for a horizontal ellipse with center $(- 2, 1), a^{2} = 16, b^{2} = 12,$ and $c^{2} = a^{2} - b^{2} = 16 - 12 = 4 .$ Thus, $a = 4, b = \sqrt{12} = 2 \sqrt{3},$ and $c = 2 .$

From the table on page 680, we have the following information:

The length of the major axis is $2 a = 8 .$

The length of the minor axis is $2 b = 4 \sqrt{3} .$

$\begin{array}{lrcl} Center: & (h, k) & = & (- 2, 1) \\ Foci: & (h \pm c, k) & = & (- 2 \pm 2, 1) \\ = & (- 4, 1) and (0, 1) \end{array}$

$\begin{array}{lrcl} Vertices: & (h \pm a, k) & = & (- 2 \pm 4, 1) \\ = & (- 6, 1) and (2, 1) \\ Endpoints of \\ minor axis: & (h, k \pm b) & = & (- 2, 1 \pm 2 \sqrt{3}) \\ = & (- 2, 1 + 2 \sqrt{3}) and (- 2, 1 - 2 \sqrt{3}) \\ \approx & (- 2, 4.46) and (- 2, - 2.46) \end{array}$

The graph of the ellipse is shown in Figure 7.23.

Figure 7.23

Ellipse with equation $3 x^{2} + 4 y^{2} + 12 x - 8 y - 32 = 0$

Practice Problem 4

Find the center, vertices, and foci of the ellipse with equation

$x^{2} + 4 y^{2} - 6 x + 8 y - 29 = 0 .$

A circle is considered an ellipse in which the two foci coincide at the center. Conversely, if the two foci of an ellipse coincide, then $2 c = 0$ (remember, 2c is the distance between the two foci of an ellipse); so $c = 0 .$ Again, from the relationship $c^{2} = a^{2} - b^{2} = 0,$ we have $a = b .$ Thus, the equation for the ellipse becomes the equation for a circle of radius $r = a,$ having the same center as that of the ellipse.

Example 5 Find an Equation for Each Ellipse Whose Graph Is Shown

Solution

The center $(h, k) = (- 2, 1)$ , so $h = - 2$ and $k = 1$ .

The distance, a, between the center $(- 2, 1)$ and the vertex $(- 2, 4)$ is $\sqrt{{(- 2 - (- 2))}^{2} + {(1 - 4)}^{2}} = \sqrt{0^{2} + 3^{2}} = \sqrt{9} = 3.$ So $a = 3$ .

The distance between the center $(- 2, 1)$ and $(- 3, 1)$ is $\sqrt{{(- 2 - (- 3))}^{2} + {(1 - 1)}^{2}} = \sqrt{1^{2} + 0^{2}} = \sqrt{1} = 1$ . So $b = 1$ .

$\begin{aligned} \frac{{(x - h)}^{2}}{b^{2}} + \frac{{(y - k)}^{2}}{a^{2}} = 1 & Equation for a vertical ellipse \\ \frac{{(x - [- 2])}^{2}}{1^{2}} + \frac{{(y - 1)}^{2}}{3^{2}} = 1 & Replace h with - 2, b with 1, k with 1, and a with 3. \\ {(x + 2)}^{2} + \frac{{(y - 1)}^{2}}{9} = 1 & Simplify . \end{aligned}$
The center $(h, k) = (0, 1),$ so $h = 0$ and $k = 1$ .

The distance, c, between the center (0, 1) and the focus at $(\sqrt{7}, 1)$ is $\sqrt{{(0 - \sqrt{7})}^{2} + {(1 - 1)}^{2}} = \sqrt{{(- \sqrt{7})}^{2} + 0^{2}} = \sqrt{7} .$ So, $c = \sqrt{7} .$

From the graph we see that the focus at $(\sqrt{7,} 1)$ and the center at (0, 1) are on the horizontal line $y = 1$ , so the ellipse is horizontal. Then $(0, - 2)$ is an endpoint of the minor axis. The distance, b, between the center (0, 1) and the endpoint $(0, - 2)$ is

$\sqrt{{(0 - 0)}^{2} + {(1 - (- 2))}^{2}} = \sqrt{0^{2} + 3^{2}} = \sqrt{9} = 3 . So b = 3 .$

$\begin{array}{l} c^{2} = a^{2} - b^{2} & Relationship between a, b, and c \\ {(\sqrt{7})}^{2} = a^{2} - 3^{2} & Replace c with \sqrt{7} and b with 3. \\ a^{2} = 7 + 9 = 16 & Solve for a^{2} . \\ \frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1 & Equation for a horizontal ellipse \\ \frac{{(x - 0)}^{2}}{16} + \frac{{(y - 1)}^{2}}{3^{2}} = 1 & Replace h with 0, a^{2} with 16, k with 1, and b with 3. \\ \frac{x^{2}}{16} + \frac{{(y - 1)}^{2}}{9} = 1 & Simplify . \end{array}$

Practice Problem 5

Find an equation for each ellipse whose graph is shown.

Applications

4 Use ellipses in applications.

Ellipses have many applications. We mention just a few:

The orbits of the planets are ellipses with the sun at one focus. This fact alone is sufficient to explain the overwhelming importance of ellipses since the seventeenth century, when Kepler and Newton did their monumental work in astronomy.
Newton also reasoned that comets move in elliptical orbits about the sun. His friend Edmund Halley used this information to predict that a certain comet (now called Halley’s comet) reappears about every 77 years. Halley saw the comet in 1682 and correctly predicted its return in 1759. The last sighting of the comet was in 1986, and it is due to return in 2061. A recent study shows that Halley’s comet has made approximately 2000 cycles so far and has about the same number to go before the sun erodes it away completely. (See Exercise 80.)
We can calculate the distance traveled by a planet in one orbit around the sun.
The reflecting property of ellipses. The reflecting property for an ellipse says that a ray of light originating at one focus will be reflected to the other focus. See Figure 7.24. Sound waves also follow such paths. This property is used in the construction of “whispering galleries,” such as the gallery at St. Paul’s Cathedral in London. Such rooms have ceilings whose cross sections are elliptical with common foci. As a result, sounds emanating from one focus are reflected by the ceiling to the other focus. Thus, a whisper at one focus may not be audible at all at a nearby place but may nevertheless be clearly heard far off at the other focus.

Example 6 illustrates how the reflecting property of an ellipse is used in lithotripsy. As the introduction to this section noted, a lithotripter is a machine that is designed to crush kidney stones into small sandlike pieces with the help of high-energy shock waves. To focus these shock waves accurately, a lithotripter uses the reflective property of ellipses. A patient is carefully positioned so that the kidney stone is at one focus of the ellipsoid (a three-dimensional ellipse) and the shock-producing source is placed at the other focus. The high-energy shock waves generated at this focus are concentrated on the kidney stone at the other focus, thus pulverizing it without harming the patient.

Example 6 Lithotripsy

An elliptical water tank has a major axis of length 6 feet and a minor axis of length 4 feet. The source of high-energy shock waves from a lithotripter is placed at one focus of the tank. To smash the kidney stone of a patient, how far should the stone be positioned from the source?

Solution

Since the length of the major axis of the ellipse is 6 feet, we have $2 a = 6;$ so $a = 3 .$ Similarly, the minor axis of 4 feet gives $2 b = 4,$ or $b = 2 .$ To find c, we use the equation $c^{2} = a^{2} - b^{2} .$ We have

$c^{2} = {(3)}^{2} - {(2)}^{2} = 5 . Therefore, c = \pm \sqrt{5} .$

If we position the center of the ellipse at (0, 0) and the major axis along the x-axis, then the foci of the ellipse are $(- \sqrt{5}, 0)$ and $(\sqrt{5}, 0) .$ The distance between these foci is $2 \sqrt{5} \approx 4.472$ feet. The kidney stone should be positioned 4.472 feet from the source of the shock waves.

Practice Problem 6

In Example 6 , if the tank has a major axis of 8 feet and a minor axis of 4 feet, how far should the stone be positioned from the source?

Section 7.3 Exercises

Concepts and Vocabulary

An ellipse is the set of all points in the plane, the of whose distances from two fixed points is a constant.
The points of intersection of the ellipse with the line through the foci are called of the ellipse.
The standard equation of an ellipse with center (0, 0), vertices $(\pm a, 0),$ and foci $(\pm c, 0)$ is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ where $b^{2} =$ .
In the equation $\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1,$ if (i) $m^{2} > n^{2},$ the graph is a(n) ellipse; (ii) if $m^{2} < n^{2},$ the graph is a(n) ellipse; (iii) if $m^{2} = n^{2},$ the graph is a(n) .
True or False. The ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ lies inside the box formed by the four lines $x = \pm a, y = \pm b .$
True or False. The graph of $A x^{2} + C y^{2} + D x + E y + F = 0$ is a degenerate conic or an ellipse if $A C > 0 .$
True or False. If the vertices of an ellipse are $(- 3, 2)$ and (5, 2), then the center of the ellipse is (1, 2).
True or False. If the graph of the equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is a circle and $a = 5,$ then $b = - 5$ .

Building Skills

In Exercises 9–22, find the standard form of the equation for the ellipse satisfying the given conditions. Graph the equation.

Foci: $(\pm 1, 0);$ vertex (3, 0)
Foci: $(\pm 3, 0);$ vertex (5, 0)
Foci: $(0, \pm 2);$ vertex (0, 4)
Foci: $(0, \pm 3);$ vertex $(0, - 6)$
Foci: $(\pm 4, 0);$ y-intercepts: $\pm 3$
Foci: $(\pm 3, 0);$ y-intercepts: $\pm 4$
Foci: $(0, \pm 2);$ x-intercepts: $\pm 4$
Foci: $(0, \pm 3);$ x-intercepts: $\pm 5$
Center: (0, 0), vertical major axis of length 10, and minor axis of length 6
Center: (0, 0), vertical major axis of length 8, and minor axis of length 4
Center: (0, 0); vertices: $(\pm 6, 0); c = 3$
Center: (0, 0); vertices: $(0, \pm 5); c = 3$
Center: (0, 0); foci: $(0, \pm 2); b = 3$
Center: (0, 0); foci: $(\pm 3, 0); b = 2$

In Exercises 23–42, find the vertices and foci for each ellipse. Graph each equation.

$\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1$
$\frac{x^{2}}{4} + \frac{y^{2}}{16} = 1$
$\frac{x^{2}}{9} + y^{2} = 1$
$x^{2} + \frac{y^{2}}{4} = 1$
$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$
$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
$\frac{x^{2}}{16} + \frac{y^{2}}{36} = 1$
$\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$
$x^{2} + y^{2} = 4$
$x^{2} + y^{2} = 16$
$x^{2} + 4 y^{2} = 4$
$9 x^{2} + y^{2} = 9$
$9 x^{2} + 4 y^{2} = 36$
$4 x^{2} + 25 y^{2} = 100$
$3 x^{2} + 4 y^{2} = 12$
$4 x^{2} + 5 y^{2} = 20$
$5 x^{2} - 10 = - 2 y^{2}$
$3 y^{2} = 21 - 7 x^{2}$
$2 x^{2} + 3 y^{2} = 7$
$3 x^{2} + 4 y^{2} = 11$

In Exercises 43 and 44, find the vertices and the foci of each ellipse.

Vertical ellipse with center at $(- 2, 3),$ major axis of length 10 and minor axis of length 6.
Horizontal ellipse with center at $(1, - 2),$ major axis of length 8 and minor axis of length 4.

In Exercises 45–50, find an equation of the ellipse satisfying the given conditions.

Vertices at (1, 0) and (9, 0) and a focus at (7, 0)
Vertices at (0, 10) and $(0, - 2)$ and a focus at (0, 7)
Vertices at $(2, - 5)$ and (2, 3) and a minor axis of length 2
Vertices at $(- 1, 4)$ and (7, 4) and a minor axis of length 1
Foci at $(- 2, 3)$ and (4, 3) and a vertex at $(- 5, 3)$
Foci at (5, 1) and (13, 1) and a vertex at $(- 3, 1)$

In Exercises 51–62, find the center, foci, and vertices of each ellipse. Graph each equation.

$\frac{{(x - 1)}^{2}}{4} + \frac{{(y - 1)}^{2}}{9} = 1$
$\frac{x^{2}}{16} + \frac{{(y + 3)}^{2}}{4} = 1$
$4 {(x + 3)}^{2} + 5 {(y - 1)}^{2} = 20$
$25 {(x - 2)}^{2} + 4 {(y + 5)}^{2} = 100$
$5 x^{2} + 9 y^{2} + 10 x - 36 y - 4 = 0$
$9 x^{2} + 4 y^{2} + 72 x - 24 y + 144 = 0$
$9 x^{2} + 5 y^{2} + 36 x - 40 y + 71 = 0$
$3 x^{2} + 4 y^{2} + 12 x - 16 y - 32 = 0$
$x^{2} + 2 y^{2} - 2 x + 4 y + 1 = 0$
$2 x^{2} + 2 y^{2} - 12 x - 8 y + 3 = 0$
$2 x^{2} + 9 y^{2} - 4 x + 18 y + 12 = 0$
$3 x^{2} + 2 y^{2} + 12 x - 4 y + 15 = 0$

In Exercises 63–68, find an equation for the ellipse whose graph is shown.

Applying the Concepts

Semielliptical arch. A portion of an arch in the shape of a semiellipse (half of an ellipse) is 50 feet wide at the base and has a maximum height of 20 feet. Find the height of the arch at a distance of 10 feet from the center. (See the figure.)
Semielliptical arch. In Exercise 69, find the distance between the two points at the base where the height of the arch is 10 feet.
Semielliptical arch bridge. A bridge is built in the shape of a semielliptical arch. The span of the bridge is 150 meters, and its maximum height is 45 meters. There are two vertical supports, each at a distance of 25 meters from the central position. Find their heights.
Elliptical billiard table. Some billiard tables (similar to pool tables, but without pockets) are manufactured in the shape of an ellipse. The foci of such tables are plainly marked for the convenience of the players. An elliptical billiard table is 6 feet long and 4 feet wide. Find the location of the foci.
Whispering gallery. The vertical cross section of the dome of Statuary Hall in Washington, D.C., is semielliptical. The hall is 96 feet long and 23 feet high. It is said that John C. Calhoun used the whispering-gallery phenomenon to eavesdrop on his adversaries. Where should Calhoun and his foes be standing for the maximum whispering effect?
Whispering gallery. The vertical cross section of the dome of the Mormon Tabernacle in Salt Lake City, Utah, is semielliptical in shape. The cross section is 250 feet long with a maximum height of 80 feet. Where should two people stand to maximize the whispering effect?

In Exercises 75–78, use the fact (discovered by Johannes Kepler in 1609) that the planets move in elliptical orbits with the sun at one of the foci. Astronomers have measured the perihelion (the smallest distance from the planet to the sun) and aphelion (the largest distance from the planet to the sun) for each of the planets. The distances given in Table 7.1 are in millions of miles.

Table 7.1

Planet	Perihelion	Aphelion
Mercury	28.56	43.88
Venus	66.74	67.68
Earth	91.38	94.54
Mars	128.49	154.83
Jupiter	460.43	506.87
Saturn	837.05	936.37
Uranus	1699.45	1866.59
Neptune	2771.72	2816.42

Mars. Find an equation of Mars’s orbit about the sun.

[Hint: Set up a coordinate system with the sun at one focus and the major axis lying on the x-axis. (See the figure.) Calculate a from the equation $2 a = aphelion + perihelion .$ Calculate c from the equation $c = a - perihelion .$ Calculate $b^{2}$ from the equation $b^{2} = a^{2} - c^{2} .$ Write the equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 .$ ]
Earth. Write an equation for the orbit of Earth about the sun.
Mercury. Write an equation for the orbit of the planet Mercury about the sun.
Saturn. Write an equation for the orbit of the planet Saturn about the sun.
Moon. The moon orbits Earth in an elliptical path with Earth at one focus. The major and minor axes of the orbit have lengths of 768,806 kilometers and 767,746 kilometers, respectively. Find the perihelion and aphelion of the orbit.
Halley’s comet. In 1682, Edmund Halley (1656–1742) identified the orbit of a certain comet (now called Halley’s comet) as an elliptical orbit about the sun, with the sun at one focus. He calculated the length of its major axis to be approximately $5.39 \times 10^{9}$ kilometers and its minor axis to be approximately $1.36 \times 10^{9}$ kilometers. Find the perihelion and aphelion of the orbit of Halley’s comet.
Orbit of a satellite. Assume that Earth is a sphere with a radius of 3960 miles. A satellite has an elliptical orbit around Earth with the center of Earth as one of its foci. The maximum and minimum heights of the satellite from the surface of Earth are 164 miles and 110 miles, respectively. Find an equation for the orbit of the satellite.

Beyond the Basics

Find an equation of the ellipse in standard form with center (0, 0) and with major axis of length 10, passing through the point $(- 3, \frac{16}{5}) .$
Find an equation of the ellipse in standard form with center (0, 0) and with major axis of length 6, passing through the point $(1, \frac{3 \sqrt{5}}{2}) .$
Find the lengths of the major and minor axes of the ellipse

$\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1,$

given that the ellipse passes through the points (2, 1) and $(1, - 3) .$
Find an equation of the ellipse in standard form with center (2, 1), passing through the points (10, 1) and (6, 2).
The eccentricity of an ellipse, denoted by e, is defined as $e = \frac{Distance between the foci}{Distance between the vertices} = \frac{2 c}{2 a} = \frac{c}{a} .$ What happens when $e = 0$ ?

In Exercises 87–91, use the definition of e given in Exercise 86 to determine the eccentricity of each ellipse. (This e has nothing to do with the number e, discussed in Chapter 4, that is the base for natural logarithms.)

$\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$
$20 x^{2} + 36 y^{2} = 720$
$x^{2} + 4 y^{2} = 1$
$\frac{{(x + 1)}^{2}}{25} + \frac{{(y - 2)}^{2}}{9} = 1$
$x^{2} + 2 y^{2} - 2 x + 4 y + 1 = 0$
An ellipse has its major axis along the x-axis and its minor axis along the y-axis. Its eccentricity is $\frac{1}{2},$ and the distance between the foci is 4. Find an equation of the ellipse.
Find an equation of the ellipse whose major axis has endpoints $(- 3, 0)$ and (3, 0) and that has eccentricity $\frac{1}{3} .$
A point P(x, y) moves so that its distance from the point (4, 0) is always one-half its distance from the line $x = 16 .$ Show that the equation of the path of P is an ellipse of eccentricity $\frac{1}{2} .$

In Exercises 95–98, find all points of intersection of the given curves by solving systems of nonlinear equations. Sketch the graphs of the curves and show the points of intersection.

${\begin{array}{rclcr} x & + & 3 y & = & - 2 \\ 4 x^{2} & + & 3 y^{2} & = & 7 \end{array}$
${\begin{array}{rcrcl} x^{2} & = & 2 y \\ 2 x^{2} & + & y^{2} & = & 12 \end{array}$
${\begin{array}{r} x^{2} + y^{2} = 20 \\ 9 x^{2} + y^{2} = 36 \end{array}$
${\begin{array}{rcrcl} 9 x^{2} & + & 16 y^{2} & = & 36 \\ 18 x^{2} & + & 5 y^{2} & = & 45 \end{array}$

In Exercises 99 and 100, use the following definition. A tangent line to an ellipse is a line that intersects the ellipse at exactly one point.

Find the equation of the tangent line to the ellipse with equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ with $a = \frac{5}{2}$ and $b = 5$ at the point (2, 3).

[Hint: See the steps for Exercise 91 in Section 7.2 and (in Step 3) write the discriminant as a perfect square.]
Show that the equation of the tangent line to the ellipse with equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ at the point $(x_{1}, y_{1}),$ is $y = - \frac{b^{2} x_{1}}{a^{2} y_{1}} x + \frac{b^{2} x_{1}^{2}}{b^{2} y_{1}^{2}} + y_{1} .$

[Hint: See the steps for Exercise 91 in Section 5.2 and show that ${(a^{2} y_{1} m + b^{2} x_{1})}^{2} = 0 .$ ]

In Exercises 101 and 102, use the fact that the area inside an ellipse is $π a b,$ where a and b are the lengths of the semi-major axis and semi-minor axis, respectively. Find the area inside each ellipse.

$9 {(x + 1)}^{2} + 16 {(y - 2)}^{2} = 144$
$5 x^{2} + 5 y^{2} + 10 x - 20 y - 62 = 0$

In Exercises 103 and 104, find the area between the two given ellipses.

$16 {(x - 1)}^{2} + 9 {(y + 2)}^{2} = 144$ , and $16 {(x - 1)}^{2} + 4 {(y + 2)}^{2} = 64$ .
$x^{2} + y^{2} + 4 x + 2 y - 20 = 0$ , and $16 x^{2} + 25 y^{2} + 64 x + 50 y - 311 = 0$ .

In Exercises 105–112, use the following definition: A latus rectum of an ellipse is a line segment that is perpendicular to the major axis passing through either of the foci and whose endpoints lie on the ellipse.

Show that the length of a latus rectum of an ellipse with equation

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, or \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1, is \frac{2 b^{2}}{a} \cdot$

In Exercises 106–109, find the length of a latus rectum of each ellipse.

$\frac{x^{2}}{25} + \frac{y^{2}}{36} = 1$
$\frac{x^{2}}{16} + \frac{y^{2}}{8} = 1$
$9 x^{2} + 4 y^{2} = 36$
$5 x^{2} + 9 y^{2} = 45$
Find an equation (or equations) of the ellipse with center (0, 0), latus rectum of length $\frac{50}{13}$ and minor axis of length 10.
Find an equation of the ellipse with center (0, 0), major axis of length 3 times the length of the minor axis, and latus rectum of length 2.
Find an equation of the ellipse with center (0, 0), eccentricity $(e = \frac{c}{a}) \frac{5}{8}$ , and latus rectum length $\frac{39}{4}$ .

Critical Thinking/Discussion/Writing

Find the number of circles with a radius of three units that touch both of the coordinate axes. Write the equation of each circle.
Find the number of ellipses with a major axis and a minor axis of lengths six units and four units, respectively, that touch both axes. Write the equation of each ellipse.

Getting Ready for the Next Section

In Exercises 115 and 116, find the slope–intercept form of the line with the given properties.

Slope: $- 1$ ; passing through $(- 3, 5)$ .
Passing through $(1, - 1)$ perpendicular to the line with equation $y = - \frac{1}{2} x + 3$ .

In Exercises 117 and 118, find the point of intersection, if any.

${\begin{cases} y = 3 x + 6 \\ y = - 3 x \end{cases}$
${\begin{cases} y = \frac{3}{4} x - 2 \\ y = - \frac{3}{4} x + 3 \end{cases}$

In Exercises 119 and 120, find the x- and y-intercepts of the graph of the given equation, if any.

$4 y^{2} - x^{2} = 1$
$16 x^{2} - 9 y^{2} = 144$

In Exercises 121 and 122, solve the equation $c^{2} = a^{2} + b^{2}$ for the specified variable.

Solve for a, when $c = \sqrt{29}$ and $b = 4 .$
Solve for b, when $a = 3$ and $c = 12 .$

In Exercises 123 and 124, find the asymptotes of the graph of the given function.

$f (x) = \frac{x^{2} + 1}{x - 1}$
$f (x) = \frac{x^{2} - x}{x + 1}$

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Table of Contents for Section 7.3 The Ellipse

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Table of Contents for
Section 7.3 The Ellipse