JEE Main 2014 · 9 Apr · Shift 1 · Mathematics Q82

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MathematicsSingle correct+4 / -1Question 82 of 90

If OB is the semi-minor axis of an ellipse, $F_1$ and $F_2$ are its foci and the angle between $F_1B$ and $F_2B$ is a right angle, then the square of the eccentricity of the ellipse is :

\dfrac12

\dfrac1{\sqrt2}

\dfrac1{2\sqrt2}

\dfrac14

Answer and worked solution

Official answer: A

$B=(0,b)$ and $F_{1,2}=(\pm c,0)$ . $\angle F_1BF_2=90^\circ$ means $b=c$ . Then $a^2=b^2+c^2=2c^2$ and $e^2=\frac{c^2}{a^2}=\frac12$ .

Topics:Ellipse

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JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 82