Question

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 $\frac{x^2}{9}+\frac{y^2}{8}=1.$ 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 tha 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

The correct option is C $5:8$


Equation of tangent at point $M(\frac{3}{2},\sqrt{6})$ to the ellipse is $\frac{x\left(\frac{3}{2}\right)}{9}+\frac{y\sqrt{6}}{8}=1$
Put $y=0\Rightarrow R\text{is}(6,0)[\because R$ lies on $x-$ axis)

Equation of the normal to the parabola at point $M(\frac{3}{2},\sqrt{6})$ is
$y-\sqrt{6}=\frac{-\sqrt{6}}{2}(x-\frac{3}{2})$
Put $y=0\Rightarrow Q\text{is}(\frac{7}{2},0)[\because Q$ lies on $x-$ axis)

$\text{Area of}△MQR=\frac{1}{2}\times \sqrt{6}\times \frac{5}{2}=\frac{5\sqrt{6}}{4}$

$\text{Area of quadrilateral}M{F}_{1}N{F}_2=\text{Area of}△M{F}_1{F}_2+\text{Area of}△N{F}_1{F}_2=\sqrt{6}+\sqrt{6}=2\sqrt{6}$
$\therefore \text{Area of}△MQR:\text{Area of quadrilateral}M{F}_{1}N{F}_2=5:8$