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 MQR to area of the quadriteral $M{F}_{1}N{F}_2$ 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 is: $\frac{x\times 3}{2\times 9}+\frac{y\sqrt{6}}{8}=1$

Let us put $y=0$ as the intersection will be on X-axis.

$\therefore$ R$=(6,0)$

Equation of normal at m is: $\sqrt{\frac{3}{2}}x+y=2\sqrt{\frac{3}{2}}+{\left(\sqrt{\frac{3}{2}}\right)}^3$

Putting $y=0$, we get: $x=2+\frac{3}{2}=\frac{7}{2}$

$\therefore$ Q$=(\frac{7}{2},0)$

$\therefore$ Area ($\Delta$MQR)$=\frac{1}{2}\times (6-\frac{7}{2})\times \sqrt{6}=\frac{5}{4}\sqrt{6}$ sq. unit

Area of quadrilateral (MF${}_1$NF${}_2$)$=2\times$ Area ($\Delta$ F${}_1$F${}_2$M)$=2\times \frac{1}{2}\times 2\times \sqrt{6}=2\sqrt{6}$ sq. unit

$\therefore$ required ratio $=\frac{\frac{5}{4}}{2}=\frac{5}{8}$