JEE Advanced 2016 · Paper 2 · Mathematics Q54

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MathematicsSingle correct+3Question 54 of 54

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PARAGRAPH 2

Let $F_1(x_1,0)$ and $F_2(x_2,0)$ , for $x_1<0$ and $x_2>0$ , be the foci of the ellipse $\dfrac{x^2}9+\dfrac{y^2}8=1$ . Suppose a parabola having vertex at the origin and focus at $F_2$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.

If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the $x$ -axis at $Q$ , then the ratio of area of the triangle $MQR$ to area of the quadrilateral $MF_1NF_2$ is

3 : 4

4 : 5

5 : 8

2 : 3

Answer and worked solution

Official answer: C

Point R. The tangent to the ellipse at M is $\dfrac{x}{6}+\dfrac{\sqrt6\,y}8=1$ . By symmetry R is on the $x$ -axis: $R=(6,0)$ .

Point Q. For $y^2=4x$ , M has parameter $t=\frac{\sqrt6}2$ . The normal at M meets the $x$ -axis at $x=2+t^2=\frac72$ .

Areas.

$[MQR]=\frac12\cdot\frac52\cdot\sqrt6=\dfrac{5\sqrt6}4$ .
$MF_1NF_2$ is a kite with diagonals $F_1F_2=2$ and $MN=2\sqrt6$ , so its area is $2\sqrt6$ .

Ratio. $\dfrac{5\sqrt6/4}{2\sqrt6}=\dfrac58$ .

(C)

Topics:Tangents and normals to conics Ellipse Parabola

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JEE Advanced 2016 · Paper 2, Mathematics question 54