1. Tardigrade
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  4. 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 ( x 2/9)+( y 2/8)=1 . Suppose a parabola having vertex at the origin and focus at F2 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 1 NF 2 is

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Q. Let $F _{1}\left( x _{1}, 0\right)$ and $F _{2}\left( x _{2}, 0\right)$, for $x _{1}<0$ and $x _{2}>0$, be the foci of the ellipse $\frac{ x ^{2}}{9}+\frac{ y ^{2}}{8}=1 .$ 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 $MF _{1} NF _{2}$ is

JEE AdvancedJEE Advanced 2016

Solution:

Let tangents to ellipse at $M$ and $N$ meet at $R \left( x _{1}, y _{1}\right)$
Then chord of contact will be
$\frac{x x_{1}}{9}+\frac{y y_{1}}{8}-1=0$
comparing it with $x=\frac{3}{2}$
$x _{1}=6, y _{1}=0 \,\,\,\therefore R (6,0)$
Normal at $M$ to the parabola intersects $x$-axis at $Q \left( x _{2}, 0\right)$
So $\frac{0-\sqrt{6}}{x_{2}-3 / 2}=m=\frac{-\sqrt{6}}{2} \,\,\,(m=$ slope of normal)
$\Rightarrow x_{3}=\frac{7}{2}$
$\therefore Q\left(\frac{7}{2}, 0\right)$
Area of $\Delta MOR =\Delta_{1}=\frac{1}{2}\begin{vmatrix}1 & 1 & 1 \\ \frac{3}{2} & \frac{7}{2} & 6 \\ \sqrt{6} & 0 & 0\end{vmatrix}=\frac{5 \sqrt{6}}{4}$
Area of $MF _{1} NF _{2}=\Delta_{2}=2 \times \frac{1}{2} \times 2 \times \sqrt{6}=2 \sqrt{6}$
$\therefore \frac{\Delta_{1}}{\Delta_{2}}=\frac{5}{8}$