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