Question

If locus of feet of perpendiculars drawn from focus of parabola $y^2-6x+36=0$ upon any tangent on it, is directrix for ellipse $\frac{x^2}{a^2}+\frac{y^2}{9}=1$, then length of latus rectum of ellipse is

• A. $\frac{3}{\sqrt{2}}$
• B. $3\sqrt{2}$
• C. $6\sqrt{2}$
• D. $9$

Answer 1

Answer: (B)

$\because$ Parabola is $y^2=6(x-6)$
locus of feet of perpendicular from focus is tangent at vertex
i.e. $x=6$
For ellipse it is directrix
$\therefore \frac{a}{e}=6$
$\Rightarrow a=6e$
$\because b^2=a^2\left(1-e^2\right)$
$\Rightarrow 9=36e^2\left(1-e^2\right)$
$\Rightarrow 1=4e^2\left(1-e^2\right)$
$\Rightarrow 4e^4-4e^2+1=0$
$\Rightarrow {\left(2e^2-1\right)}^2=0$
$\Rightarrow e^2=\frac{1}{2}$
$\Rightarrow e=\frac{1}{\sqrt{2}}$
$\therefore a=3\sqrt{2}$
$\therefore$ Latus rectum $=\frac{2b^2}{a}=\frac{2\cdot 9}{3\sqrt{2}}$
$=3\sqrt{2}$