Question

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 $M{F}_{1}N{F}_2$ is

• A. $3:4$
• B. $4:5$
• C. $5:8$
• D. $2:3$

Answer 1

Answer: (C)

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 $\Delta MOR={\Delta }_1=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ \frac{3}{2}& \frac{7}{2}& 6\\ \sqrt{6}& 0& 0\end{array}\right|=\frac{5\sqrt{6}}{4}$
Area of $M{F}_{1}N{F}_2={\Delta }_2=2\times \frac{1}{2}\times 2\times \sqrt{6}=2\sqrt{6}$
$\therefore \frac{{\Delta }_1}{{\Delta }_2}=\frac{5}{8}$