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Ellipse question

2006 · Shift 0 · Q81

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2006 · Shift 0 · Q81

Ellipse question

2006 · Shift 0 · Q81

JEE MainMathematicsEllipseMCQ+4 / −1

In the ellipse, the distance between its foci is

6

and minor axis is

8

. Then its eccentricity is :

A
${3 \over 5}$
B
${1 \over 2}$
C
${4 \over 5}$
D
${1 \over {\sqrt 5 }}$

View written solutionFree

Correct answer: A

Use standard ellipse notation

For an ellipse,

distance between foci $= 2c$
length of minor axis $= 2b$
and the relation is $a^2=b^2+c^2$
eccentricity is $e=\frac{c}{a}$

Extract the given information

The distance between foci is $6$ , so $2c=6 \implies c=3$

The minor axis is $8$ , so $2b=8 \implies b=4$

Find $a$ using ellipse relation

$a^2=b^2+c^2=4^2+3^2=16+9=25$

So, $a=5$

Compute eccentricity

$e=\frac{c}{a}=\frac{3}{5}$

Match with the options

$\frac{3}{5}$ corresponds to Option A.

Therefore, the eccentricity of the ellipse is $\boxed{\frac{3}{5}}$ .

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