JEE Main 2015 · 11 Apr · Shift 1 · Mathematics Q81

MathematicsSingle correct+4 / -1Question 81 of 90

If the distance between the foci of an ellipse is half the length of its latus rectum, then the eccentricity of the ellipse is :

\dfrac12

\dfrac{2\sqrt2-1}2

\sqrt2-1

\dfrac{\sqrt2-1}2

Answer and worked solution

Official answer: C

$2ae=\frac12\cdot\frac{2b^2}a$ , so $2e=1-e^2$ . Then $e^2+2e-1=0$ and $e=\sqrt2-1$ .

JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 81