Let the hyperbola H: x^2 a^2- y^2 b^2=1 pass through the point (2√2,-2√2). A parabola is drawn whose focus is same as the focus of H with positive…

Given,

$$H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$$H:a2x2​−b2y2​=1

So coordinates of foci will be : $S(ae,0),{S}^{'}(-ae,0)$S(ae,0),S′(−ae,0)

Now foot of directrix of parabola will be $(-ae,0)$(−ae,0)

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

Now, semi latus rectum of parabola $=|S{S}^{'}|=2ae$=∣SS′∣=2ae

Given, $4ae=e(\frac{2{b}^{2}}{a})$4ae=e(a2b2​)

$$\Rightarrow {b}^{2}=2{a}^{2}...(1)$$⇒b2=2a2...(1)

Also given, $(2\sqrt{2},-2\sqrt{2})$(22​,−22​) lies on $H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$H:a2x2​−b2y2​=1

$$\Rightarrow \frac{{(2\sqrt{2})}^{2}}{{a}^{2}}-\frac{{(2\sqrt{2})}^{2}}{{b}^{2}}=1$$⇒a2(22​)2​−b2(22​)2​=1

$$\Rightarrow \frac{1}{{a}^{2}}-\frac{1}{{b}^{2}}=\frac{1}{8}...(2)$$⇒a21​−b21​=81​...(2)

Now from equation (1)&(2) we get,

$${a}^{2}=4,{b}^{2}=8$$a2=4,b2=8

$$\because {b}^{2}={a}^{2}({e}^{2}-1)$$∵b2=a2(e2−1)

$$\therefore e=\sqrt{3}$$∴e=3​

So, the equation of parabola is ${y}^{2}=4\times (ae)x$y2=4×(ae)x$\Rightarrow {y}^{2}=8\sqrt{3}x$⇒y2=83​x

So, only $(3\sqrt{3},-6\sqrt{2})$(33​,−62​) will satisfy the parabola ${y}^{2}=8\sqrt{3}x$y2=83​x