Question

Let $F_1(x_1,0)$ and $F_2(x_2,0),$ where, $x_1<0$ 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 $\Delta MQR$ 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 $M(\frac{3}{2},\sqrt{6})$ to $\frac{x^2}{9}+\frac{y^2}{8}=1$ is
$\frac{3}{2}.\frac{x}{9}+\sqrt{6}\frac{y}{8}=1$
Which intersect X-axis at(6,0).
Also equation of tangent at N $(\frac{3}{2},-\sqrt{6})$ is
$\frac{3}{2}.\frac{x}{9}-\sqrt{6}\frac{y}{8}=1$
Both the tangents intersects x-Axis at R (6,0)
Also normal to the parabola at M $(\frac{3}{2},\sqrt{6})$ is $y-\sqrt{6}=\frac{-\sqrt{6}}{2}(x-\frac{3}{2})$
On solving with $y=0$, we get $(\frac{7}{2},0)$

$\therefore areaof\Delta MQR=\frac{1}{2}(6-\frac{7}{2})\sqrt{6}=\frac{5\sqrt{6}}{4}$sq units
and area of quadrilateral $M{F}_{1}N{F}_2=2\times \left(\frac{1}{2}\right(1-(-1))\sqrt{6})=2\sqrt{6}$ sq units
$\therefore \frac{areaof\Delta MQR}{Areaofquadrilateral{M}_{1}N{F}_2}=\frac{5}{8}$