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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?
$(2 \sqrt{3}, 3 \sqrt{2})$
$(3 \sqrt{3},-6 \sqrt{2})$
$(\sqrt{3},-\sqrt{6})$
$(3 \sqrt{6}, 6 \sqrt{2})$
Solution
$H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
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 $=\left| SS ^{\prime}\right|=2 ae$
Given, $4 a e = e \left(\frac{2 b ^{2}}{ a }\right)$
$b ^{2}=2 a ^{2}$
Given, $(2 \sqrt{2},-2 \sqrt{2})$ lies on $H$
$\frac{1}{ a ^{2}}-\frac{1}{ b ^{2}}=\frac{1}{8}$
From $(1)$ and $(2)$
$a^{2}=4, b^{2}=8$
$\because b^{2}=a^{2}\left(e^{2}-1\right)$
$\therefore e=\sqrt{3}$
Equation of parabola is $y^{2}=8 \sqrt{3} x$
