# Class 11 Maths MCQ – Conic Sections – Ellipse

This set of Class 11 Maths Chapter 11 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

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 F1 and F2 are foci then |PF1+PF2| remains constant.

2. The center of ellipse is same as a vertex.
a) True
b) False

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)

Explanation: Comparing the equation with $$(\frac{x}{a})^2+(\frac{y}{b})^2$$=1, we get a=5 and b=4.
For ellipse, c2=a2-b2=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)

Explanation: Comparing the equation with $$(\frac{x}{b})^2+(\frac{y}{a})^2$$=1, we get a=5 and b=4.
For ellipse, c2=a2-b2=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

Explanation: Comparing the equation with $$(\frac{x}{a})^2+(\frac{y}{b})^2$$=1, we get a=5 and b=4.
For ellipse, c2=a2-b2= 25-16=9 => c=3.
We know, for ellipse c=a*e
So, e=c/a = 3/5.
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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

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

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

Explanation: We know, length of latus rectum = 2b2/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

Explanation: Comparing the equation with $$(\frac{x}{a})^2+(\frac{y}{b})^2$$=1, we get a=5 and b=4.
For ellipse, c2=a2-b2=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

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

Explanation: Given, a=5 and c=3 => b2=a2-c2 = 52-32=42 => 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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