Let the focal chord of the parabola P: y ^ 2 = 4x along the line L:...

Given, parabola ${y}^{2} = 4x$ and line $L: y = mx + c$,

Now plotting the diagram we get,

Now focus of hyperbola, $H: \dfrac{{x}^{2}}{4} - \dfrac{{y}^{2}}{4} = 1$ will be \left(ae, 0\right)$⇒ F\left(2\sqrt{2}, 0\right)$

Now given line $L: y = mx + c$ pass focus of parabola \left(1, 0\right)$</p><p>$⇒ 0 = m + c ⋯ \left(1\right)$</p><p>Now also given line$L$is tangent to Hyperbola.$\dfrac{{x}^{2}}{4} - \dfrac{{y}^{2}}{4} = 1$</p><p>So by using condition of tangent to hyperbola in slope form we get,$c = \pm \sqrt{{a}^{2}{m}^{2} - {l}^{2}}$</p><p>$⇒ c = \pm \sqrt{4{m}^{2} - 4}$</p><p>Now from$\left(1\right)$we get,$- m = \pm \sqrt{4{m}^{2} - 4}$</p><p>Squaring both side we get,${m}^{2} = 4{m}^{2} - 4$</p><p>$⇒ 4 = 3{m}^{2}$</p><p>$⇒ m = \dfrac{2}{\sqrt{3}}$(as$m > 0$)</p><p>So,$c = - m⇒ c = \dfrac{- 2}{\sqrt{3}}$</p><p>So equation of line will be $y = \dfrac{2x}{\sqrt{3}} - \dfrac{2}{\sqrt{3}}$</p><p>Now finding intersecting point of line and parabola ${y}^{2} = 4x$ we get,</p><p>$⇒ {\left(\dfrac{2x - 2}{\sqrt{3}}\right)}^{2} = 4x$</p><p>$⇒ {x}^{2} + 1 - 2x = 3x$</p><p>$⇒ {x}^{2} - 5x + 1 = 0$</p><p>$⇒ {x}_{1} + {x}_{2} = 5 \text{ and } {x}_{1}{x}_{2} = 1$</p><p>${y}^{2} = 4\left(\dfrac{\sqrt{3}y + 2}{2}\right)$</p><p>${y}^{2} = 2\sqrt{3}y + 4$</p><p>$⇒ {y}^{2} - 2\sqrt{3}y - 4 = 0$</p><p>$⇒ {y}_{1} + {y}_{2} = 2\sqrt{3} \text{ and } {y}_{1}{y}_{2} = - 4$</p><p>Area of required quadrilateral will be </p><p>$= \left|\dfrac{1}{2}\left|$\begin{cases}0 & {x}_{1} & 2\sqrt{2} & {x}_{2} & 0 \\ 0 & {y}_{1} & 0 & {y}_{2} & 0\end{cases}$\right|\right|$</p><p>$= \left|\dfrac{1}{2}\left[- 2\sqrt{2}{y}_{1} + 2\sqrt{2}{y}_{2}\right]\right|$</p><p>$= \sqrt{2}\left|{y}_{2} - {y}_{1}\right| = \dfrac{\left(\sqrt{2}\right)\sqrt{12 + 16}}{111}$</p><p>$= \sqrt{56}= 2\sqrt{14}