IIT-JEE 2009 · Paper 2 · Mathematics Q22

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MathematicsSingle correct+3 / -1Question 22 of 57

The normal at a point $P$ on the ellipse $x^2+4y^2=16$ meets the $x$ -axis at $Q$ . If $M$ is the mid point of the line segment $PQ$ , then the locus of $M$ intersects the latus rectums of the given ellipse at the points

\left(\pm\dfrac{3\sqrt5}2,\pm\dfrac27\right)

\left(\pm\dfrac{3\sqrt5}2,\pm\dfrac{\sqrt{19}}4\right)

\left(\pm2\sqrt3,\pm\dfrac17\right)

\left(\pm2\sqrt3,\pm\dfrac{4\sqrt3}7\right)

Answer and worked solution

Official answer: C

Normal. On $\dfrac{x^2}{16}+\dfrac{y^2}4=1$ take $P(4\cos\theta,2\sin\theta)$ . The normal $\dfrac{4x}{\cos\theta}-\dfrac{2y}{\sin\theta}=12$ meets the x-axis at $Q(3\cos\theta,0)$ .

Midpoint. $M=\left(\frac72\cos\theta,\sin\theta\right)$ , so its locus is $\dfrac{4x^2}{49}+y^2=1$ .

Latus rectum. $e=\frac{\sqrt3}2$ , so $x=\pm ae=\pm2\sqrt3$ ; then $y^2=1-\dfrac{48}{49}=\dfrac1{49}$ .

Points $\left(\pm2\sqrt3,\pm\frac17\right)$ .

(C)

Topics:Ellipse Tangents and normals to conics Locus

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