Mathematics Questions and Answers – Conic Sections – Ellipse

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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

Answer: d
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.
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2. The center of ellipse is same as a vertex.
a) True
b) False
View Answer

Answer: b
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

Answer: a
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).
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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

Answer: c
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
View Answer

Answer: b
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
View Answer

Answer: d
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

Answer: c
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.
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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

Answer: b
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
View Answer

Answer: a
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.
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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

Answer: b
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

Answer: a
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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter