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?

Given hyperbola

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

passing through the point

$$(2\sqrt2,-2\sqrt2)$$

Substituting the point in the hyperbola,

$$\frac{(2\sqrt2)^2}{a^2}-\frac{(-2\sqrt2)^2}{b^2}=1$$

$$\frac8{a^2}-\frac8{b^2}=1$$

$$8(b^2-a^2)=a^2b^2\qquad\cdots(1)$$

For the hyperbola,

$$e^2=1+\frac{b^2}{a^2}$$

Length of latus rectum of hyperbola is

$$L_H=\frac{2b^2}{a}$$

The foci are

$$(\pm ae,0)$$

The parabola has focus at

$$(ae,0)$$

and its directrix passes through the other focus

$$(-ae,0)$$

Hence, the directrix is

$$x=-ae$$

Therefore, the vertex is midpoint of the focus and directrix, i.e.,

$$(0,0)$$

and the distance of focus from vertex is

$$ae$$

Thus the parabola is

$$y^2=4(ae)x$$

Hence, length of latus rectum of parabola is

$$L_P=4(ae)$$

Given,

$$L_P=eL_H$$

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

Cancelling $$2e,$$

$$2a=\frac{b^2}{a}$$

$$b^2=2a^2\qquad\cdots(2)$$

Substituting $$b^2=2a^2$$ in (1),

$$8(2a^2-a^2)=a^2(2a^2)$$

$$8a^2=2a^4$$

$$a^2=4$$

Hence,

$$a=2,\qquad b^2=8$$

Now,

$$e^2=1+\frac{8}{4}=3$$

$$e=\sqrt3$$

Thus,

$$ae=2\sqrt3$$

Therefore, the parabola is

$$y^2=8\sqrt3\,x$$

Hence, any point satisfying

$$y^2=8\sqrt3\,x$$

lies on the parabola.

By trial and error method with options,we conclude that $$(3\sqrt{3}, -6\sqrt{2})$$ lies on the parabola