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Ellipse question

2024 · 30 Jan · Shift 1 · Q41

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Ellipse
2024 · 30 Jan · Shift 1 · Q41

Ellipse question

2024 · 30 Jan · Shift 1 · Q41

JEE MainMathematicsEllipseMCQ+4 / −1

If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :

A
$\frac{1}{\sqrt{3}}$
B
$\frac{2}{\sqrt{5}}$
C
$\frac{\sqrt{3}}{2}$
D
$\frac{\sqrt{5}}{3}$

View written solutionFree

Correct answer: B

Recall standard ellipse relations

For an ellipse,

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \qquad (a>b)

we have:

Length of major axis $=2a$
Length of minor axis $=2b$
Distance between the foci $=2c$
Eccentricity $e=\dfrac{c}{a}$

Also,

c^2=a^2-b^2

Use the given condition

The question says:

length of the minor axis is equal to half of the distance between the foci.

So,

2b=\frac{1}{2}(2c)=c

Hence,

c=2b

Substitute into the ellipse relation

Using

c^2=a^2-b^2

and $c=2b$ , we get

(2b)^2=a^2-b^2

4b^2=a^2-b^2

a^2=5b^2

Therefore,

a=\sqrt{5}\,b

Find eccentricity

Since

e=\frac{c}{a}

and $c=2b$ , $a=\sqrt{5}b$ , we get

e=\frac{2b}{\sqrt{5}b}=\frac{2}{\sqrt{5}}

Check options

The eccentricity is

\boxed{\frac{2}{\sqrt{5}}}

which matches Option B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So the stored answer is correct.

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