Paragraph: 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.
Question: 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 $M Q R$ to area of the quadrilateral $M F_{1} N F_{2}$ is

• A

  $3 : 4$
• B

  $4 : 5$
• Correct answer
  C

  $5 : 8$
• D

  $2 : 3$

Solution

Equation of chord of contact MN is $x=\frac{3}{2}$
Let R(h,k)
$\Rightarrow$ Equation of chord of contact
$\Rightarrow \frac{xh}{9}+\frac{yk}{8}=1$
Comparing we get R(6,0)
Normal to parabola at $M :y- \sqrt{6}= -\frac{\sqrt{6}}{2·1} \left(x-\frac{3}{2}\right)$
Solving it with $y=0,$ we get $Q \equiv \left(\frac{7}{2}, 0\right)$
$\therefore$ Area of triangle MQR $=\frac{1}{2}. \left(6-\frac{7}{2}\right). \sqrt{6}=\frac{5\sqrt{6}}{4}$
Area of quadrilateral $M{F}_{1}N{F}_2=2·\frac{1}{2}\left(1-\left(-1\right)\right)\sqrt{6}=2\sqrt{6}$
Required ratio $= 5 : 8$

Hence, the correct answer is option: 'C'.