Let the hyperbola $H:\frac{x^2}{a^2}-\frac{y^2}{ b^2}=1$ pass through the point $\left(2\sqrt{2},-2\sqrt{2}\right)$. 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?

Given,

$H: \frac{x^2}{a^2}-\frac{y^2}{ b^2}=1$

So coordinates of foci will be : $S\left(ae,0\right),S^{\text{'}}\left(-ae,0\right)$

Now foot of directrix of parabola will be $\left(-ae,0\right)$

Also focus of parabola is which is same as focus of $H$ will be $\left(ae, 0\right)$

Now, semi latus rectum of parabola $=\left|S{S}^{\text{'}}\right|=2ae$

Given, $4ae=e\left(\frac{2b^2}{a}\right)$

$\Rightarrow b^2=2a^2 ...\left(1\right)$

Also given, $\left(2\sqrt{2},-2\sqrt{2}\right)$ lies on $H: \frac{x^2}{a^2}-\frac{y^2}{ b^2}=1$

$\Rightarrow \frac{{\left(2\sqrt{2}\right)}^2}{a^2}-\frac{{\left(2\sqrt{2}\right)}^2}{ b^2}=1$

$\Rightarrow \frac{1}{a^2}-\frac{1}{ b^2}=\frac{1}{8} ...\left(2\right)$

Now from equation $\left(1\right) \& \left(2\right)$ we get,

$a^2=4, b^2=8$

$\because b^2=a^2\left(e^2-1\right)$

$\therefore e=\sqrt{3}$

So, the equation of parabola is $y^2=4\times \left(ae\right)x$$\Rightarrow y^2=8\sqrt{3}x$

So, only $\left(3\sqrt{3},-6\sqrt{2}\right)$ will satisfy the parabola $y^2=8\sqrt{3}x$