Question

Let the hyperbola $H : \frac{ x ^2}{ a ^2}-\frac{y^2}{b^2}=1$ pass through the point $(2 \sqrt{2},-2 \sqrt{2})$. A parabola is drawn whose focus is same as the focus of $H$ with positive abscissa and the directrix of the parabola passes through the other focus of $H$. If the length of the latus rectum of the parabola is e times the length of the latus rectum of $H$, where e is the eccentricity of $H$, then which of the following points lies on the parabola?

• A. $(2 \sqrt{3}, 3 \sqrt{2})$
• B. $(3 \sqrt{3},-6 \sqrt{2})$
• C. $(\sqrt{3},-\sqrt{6})$
• D. $(3 \sqrt{6}, 6 \sqrt{2})$

Answer 1

Answer: (B)

$H : \frac{ x ^2}{ a ^2}-\frac{ y ^2}{ b ^2}=1$
Foci : S (ae, 0), S' $(-a e, 0)$
Foot of directrix of parabola is $(-a e, 0)$
Focus of parabola is (ae, 0 )
Now, semi latus rectum of parabola $=\left| SS ^{\prime}\right|=2 ae$
Given, $4 a e=e\left(\frac{2 b^2}{a}\right)$
$\Rightarrow b ^2=2 a ^2$....(1)
Given, $(2 \sqrt{2},-2 \sqrt{2})$ lies on $H$
$\Rightarrow \frac{1}{ a ^2}-\frac{1}{ b ^2}=\frac{1}{8}$...(2)
From (1) and (2)
$ a ^2=4, b ^2=8$
$ \because b ^2= a ^2\left( e ^2-1\right) $
$ \therefore e =\sqrt{3} $
$\Rightarrow \text { Equation of parabola is } y ^2=8 \sqrt{3} x$