Let the hyperbola H: x ^ 2 a ^ 2 - y ^ 2 b ^ 2 = 1 pass through the...

Given,

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

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

Now foot of directrix of parabola will be \left(- ae, 0\right)$</p><p>Also focus of parabola is which is same as focus of$H$will be \left(ae, 0\right)$

Now, semi latus rectum of parabola =\left|S{S}^{'}\right| = 2ae$</p><p>Given,$4ae = e\left(\dfrac{2{b}^{2}}{a}\right)$</p><p>$⇒ {b}^{2} = 2{a}^{2} . . . \left(1\right)$</p><p>Also given, \left(2\sqrt{2}, - 2\sqrt{2}\right)$ lies on $H: \dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1$

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

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

Now from equation \left(1\right) \text{ and } \left(2\right)$we get,</p><p>${a}^{2} = 4, {b}^{2} = 8$</p><p>$∵ {b}^{2} = {a}^{2}\left({e}^{2} - 1\right)$</p><p>$∴ e = \sqrt{3}$</p><p>So, the equation of parabola is${y}^{2} = 4 \times \left(ae\right)x$$⇒ {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$