To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)

Ellipse Worksheet

Ellipse Worksheet
  • Page 1
 1.  
What is the eccentricity of the ellipse 16x2 + 49y2 = 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.]


Correct answer : (2)
 2.  
What is the length of the semi minor axis of the ellipse 64x2 + 81y2 = 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.


Correct answer : (4)
 3.  
What is the length of the semi major axis of the ellipse 64x2 + 4y2 = 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.


Correct answer : (2)
 4.  
What is the eccentricity of the ellipse 4x2 + 64y2 = 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.]


Correct answer : (3)
 5.  
Find an equation of the ellipse with foci (0, - 4) and (0, 4) whose minor axis has a length of 8 units.
a.
32x2 + 16y2 = 1
b.
y232+x216 = 1
c.
x232-y216 = 1
d.
x232+y216 = 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


Correct answer : (2)
 6.  
What is the equation of the focal axis of the ellipse x2 + 9y2 = 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.


Correct answer : (4)
 7.  
What is the equation of the focal axis of the ellipse 64x2 + y2 = 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.


Correct answer : (2)
 8.  
What are the foci of the ellipse 4x2 + 64y2 = 256?
a.
(0, ± 60)
b.
(0, ± 68)
c.
68, 0)
d.
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).]


Correct answer : (4)
 9.  
What are the foci of the ellipse 9y2 + 25x2 = 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).]


Correct answer : (2)
 10.  
Find the vertices of the ellipse x2 + 36y2 = 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).]


Correct answer : (2)

*AP and SAT are registered trademarks of the College Board.