﻿ Ellipse Worksheet | Problems & Solutions

# Ellipse Worksheet

Ellipse Worksheet
• Page 1
1.
What is the eccentricity of the ellipse 16$x$2 + 49$y$2 = 784?
 a. 0.43 b. 0.82 c. 0.22 d. - 0.82

#### Solution:

16x2 + 49y2 = 784

x249+y216 = 1
[Divide throughout by 784.]

This is in the standard form x2a2 + y2b2 = 1 where a2 = 49, b2 = 16.

The focal axis of the ellipse is x - axis.
[Denominator of x2is larger.]

Eccentricity of the ellipse, e = a2-b2a
[Use the formula.]

= 49 - 167 = 0.82
[Simplify.]

2.
What is the length of the semi minor axis of the ellipse 64$x$2 + 81$y$2 = 5184?
 a. 9 b. 64 c. 81 d. 8

#### Solution:

64x2 + 81y2 = 5184

x281+y264 = 1
[Divide both sides by 5184.]

This equation is in the standard form x2a2+y2b2 = 1, where a2 = 81, b2 = 64.

a = 9, b = 8
[Solve for a, b.]

The focal axis of the ellipse is x - axis.
[Denominator of x2 is larger.]

So, the length of the semi minor axis of the ellipse is b = 8.

3.
What is the length of the semi major axis of the ellipse 64$x$2 + 4$y$2 = 256?
 a. 2 b. 8 c. 64 d. 68

#### Solution:

64x2 + 4y2 = 256

x24+y264 = 1 or y264+x24 = 1
[Divide both sides by 256.]

This equation is in the standard form y2a2+x2b2 = 1, where a2 = 64 and b2 = 4.

a = 8 and b = 2
[Solve for a, b.]

The focal axis of the ellipse is y - axis.
[Denominator of y2 is larger.]

So, the length of the semi major axis of the parabola is a = 8.

4.
What is the eccentricity of the ellipse 4$x$2 + 64$y$2 = 256?
 a. 3.87 b. 2.06 c. 0.97 d. 0.12

#### Solution:

4x2 + 64y2 = 256

x264+y24 = 1
[Divide both sides by 256.]

This is in the standard form x2a2+y2b2 = 1, where a2 = 64, b2 = 4

The focal axis of the ellipse is x - axis.
[Denominator of x2 is larger.]

Eccentricity of the ellipse = e = a2-b2a
[Formula.]

= 64 - 48 = 0.97
[Simplify.]

5.
Find an equation of the ellipse with foci (0, - 4) and (0, 4) whose minor axis has a length of 8 units.
 a. 32$x$2 + 16$y$2 = 1 b. $\frac{{y}^{2}}{32}+\frac{{x}^{2}}{16}$ = 1 c. $\frac{{x}^{2}}{32}-\frac{{y}^{2}}{16}$ = 1 d. $\frac{{x}^{2}}{32}+\frac{{y}^{2}}{16}$ = 1

#### Solution:

The foci of the ellipse are (0, - 4) and (0, 4), which are on the y - axis with c = 4.

The center of the ellipse is (0 + 02, - 4 + 42) = (0, 0).
[Midpoint of the line segment joining foci.]

Length of the semiminor axis is b = 8 / 2 = 4

a2 = b2 + c2 = 16 + 16 = 32
[Pythagorean relation.]

So, the standard equation of the ellipse is y232+x216 = 1

6.
What is the equation of the focal axis of the ellipse $x$2 + 9$y$2 = 9?
 a. $y$ + $x$ = 0 b. $y$ - axis c. $y$ = $x$ d. $x$ - axis

#### Solution:

x2 + 9y2 = 9

x29+y21 = 1
[Divide both sides by 9.]

This equation is in the standard form x2a2+y2b2 = 1 and since the larger number is the denominator of x2, the focal axis is the x - axis.

7.
What is the equation of the focal axis of the ellipse 64$x$2 + $y$2 = 64?
 a. $y$ = $x$ b. $y$ - axis c. $x$ - axis d. $y$ + $x$ = 0

#### Solution:

64x2 + y2 = 64

y264+x21 = 1
[Divide both sides by 64.]

This equation is in the standard form y2a2+x2b2 = 1 and since the larger number is the denominator of y2, the focal axis is the y - axis.

8.
What are the foci of the ellipse 4$x$2 + 64$y$2 = 256?
 a. (0, ± $\sqrt{60}$) b. (0, ± $\sqrt{68}$) c. (± $\sqrt{68}$, 0) d. (± $\sqrt{60}$, 0)

#### Solution:

4x2 + 64y2 = 256

x264+y24 = 1
[Divide both sides by 4.]

This equation is in the standard form x2a2+y2b2 = 1, where a2 = 64, b2 = 4.

c2 = a2 - b2 = 64 - 4 = 60
[Pythagorean relation.]

c = 60
[c > 0.]

The focal axis of the ellipse is x - axis.
[Denominator of x2 is larger.]

So, the foci of the ellipse are (± 60, 0).
[Use (± c, 0).]

9.
What are the foci of the ellipse 9$y$2 + 25$x$2 = 225?
 a. (± 3, 0) b. (0, ± 4) c. (0, ± 5) d. (± 4, 0)

#### Solution:

25x2 + 9y2 = 225

y225+x29 = 1
[Divide both sides by 225.]

This equation is in the standard form y2a2+x2b2 = 1, where a2 = 25, b2 = 9.

c2 = a2 - b2 = 25 - 9 = 16
[Pythagorean relation.]

c = 4
[c > 0.]

The focal axis of the ellipse is y - axis.
[Denominator of y2 is larger.]

So, the foci of the ellipse are (0, ± 4).
[Use (0, ± c).]

10.
Find the vertices of the ellipse $x$2 + 36$y$2 = 36.
 a. (0, ± 6) b. (± 6, 0) c. (± 36, 0) d. (0, ± 36)

#### Solution:

x2 + 36y2 = 36

x236+y21 = 1
[Divide both sides by 36.]

This equation is in the standard form x2a2+y2b2 = 1, where a2 = 36, b2 = 1.

The focal axis of the ellipse is x - axis.
[Denominator of x2 is larger.]

a = 6 and b = 1
[Solve for a, b.]

So, the vertices of the ellipse x2 + 36y2 = 36 are (± 6, 0).
[Use (± a, 0).]