This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Conic Sections – Ellipse”.

1. An ellipse has ___________ vertices and ____________ foci.

a) two, one

b) one, one

c) one, two

d) two, two

View Answer

Explanation: An ellipse has two vertices lying on each end and two foci lying inside the ellipse.

If P is a point on ellipse and F

_{1}and F

_{2}are foci then |PF

_{1}+PF

_{2}| remains constant.

2. The center of ellipse is same as a vertex.

a) True

b) False

View Answer

Explanation: No, center and vertex are different for ellipse.

Ellipse has one center and two vertices.

3. Find the coordinates of foci of ellipse \((\frac{x}{25})^2+(\frac{y}{16})^2\)=1.

a) (±3,0)

b) (±4,0)

c) (0,±3)

d) (0,±4)

View Answer

Explanation: Comparing the equation with \((\frac{x}{a})^2+(\frac{y}{b})^2\)=1, we get a=5 and b=4.

For ellipse, c

^{2}=a

^{2}-b

^{2}=25-16=9 => c=3.

So, coordinates of foci are (±c,0) i.e. (±3,0).

4. Find the coordinates of foci of ellipse \((\frac{x}{16})^2+(\frac{y}{25})^2\)=1.

a) (±3,0)

b) (±4,0)

c) (0,±3)

d) (0,±4)

View Answer

Explanation: Comparing the equation with \((\frac{x}{b})^2+(\frac{y}{a})^2\)=1, we get a=5 and b=4.

For ellipse, c

^{2}=a

^{2}-b

^{2}=25-16=9 => c=3.

So, coordinates of foci are (0,±c) i.e. (0,±3).

5. What is eccentricity for \((\frac{x}{25})^2+(\frac{y}{16})^2\)=1?

a) 2/5

b) 3/5

c) 15

d) 5/3

View Answer

Explanation: Comparing the equation with \((\frac{x}{a})^2+(\frac{y}{b})^2\)=1, we get a=5 and b=4.

For ellipse, c

^{2}=a

^{2}-b

^{2}= 25-16=9 => c=3.

We know, for ellipse c=a*e

So, e=c/a = 3/5.

6. What is major axis length for ellipse \((\frac{x}{25})^2+(\frac{y}{16})^2\)=1?

a) 5 units

b) 4 units

c) 8 units

d) 10 units

View Answer

Explanation: Comparing the equation with \((\frac{x}{a})^2+(\frac{y}{b})^2\)=1, we get a=5 and b=4.

Major axis length = 2a = 2*5 =10 units.

7. What is minor axis length for ellipse \((\frac{x}{25})^2+(\frac{y}{16})^2\)=1?

a) 5 units

b) 4 units

c) 8 units

d) 10 units

View Answer

Explanation: Comparing the equation with \((\frac{x}{a})^2+(\frac{y}{b})^2\)=1, we get a=5 and b=4.

Minor axis length = 2b = 2*4 = 8 units.

8. What is length of latus rectum for ellipse \((\frac{x}{25})^2+(\frac{y}{16})^2\)=1?

a) 25/2

b) 32/5

c) 5/32

d) 8/5

View Answer

Explanation: We know, length of latus rectum = 2b

^{2}/a.

So, length of latus rectum of given ellipse = 2*42/5 = 32/5.

9. What is equation of latus rectums of ellipse \((\frac{x}{25})^2+(\frac{y}{16})^2\)=1?

a) x=±3

b) y=±3

c) x=±2

d) y=±2

View Answer

Explanation: Comparing the equation with \((\frac{x}{a})^2+(\frac{y}{b})^2\)=1, we get a=5 and b=4.

For ellipse, c

^{2}=a

^{2}-b

^{2}=25-16=9 => c=3.

Equation of latus rectum x=±c i.e. x=±3.

10. If length of major axis is 10 and minor axis is 8 and major axis is along x-axis then find the equation of ellipse.

a) \((\frac{x}{4})^2+(\frac{y}{5})^2\)=1

b) \((\frac{x}{5})^2+(\frac{y}{4})^2\)=1

c) \((\frac{x}{10})^2+(\frac{y}{8})^2\)=1

d) \((\frac{x}{8})^2+(\frac{y}{10})^2\)=1

View Answer

Explanation: Given, 2a=10 => a=5 and 2b=8 => b=4.

Equation of ellipse with major axis along x-axis is \((\frac{x}{a})^2+(\frac{y}{b})^2\)=1.

So, equation of given ellipse is \((\frac{x}{5})^2+(\frac{y}{4})^2\)=1.

11. If foci of an ellipse are (0, ±3) and length of semimajor axis is 5 units, then find the equation of ellipse.

a) \((\frac{x}{4})^2+(\frac{y}{5})^2\)=1

b) \((\frac{x}{5})^2+(\frac{y}{4})^2\)=1

c) \((\frac{x}{10})^2+(\frac{y}{8})^2\)=1

d) \((\frac{x}{8})^2+(\frac{y}{10})^2\)=1

View Answer

Explanation: Given, a=5 and c=3 => b

^{2}=a

^{2}-c

^{2}= 5

^{2}-3

^{2}=4

^{2}=> b=4.

Equation of ellipse with major axis along y-axis is \((\frac{x}{b})^2+(\frac{y}{a})^2\)=1.

So, equation of given ellipse is \((\frac{x}{4})^2+(\frac{y}{5})^2\)=1.

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