$$H:{{{x^2}} \over 4} - {{{y^2}} \over 4} = 1$$

Focus $$(ae,0)$$

$$F\left( {2\sqrt 2 ,\,0} \right)$$

$$y = mx + c$$ passes through (1, 0)

$$0 = m + C$$ ...... (i)

L is tangent to hyperbola

$$C = \, \pm \,\sqrt {4{m^2} - 4} $$

$$ - m = \, \pm \,\sqrt {4{m^2} - 4} $$

$${m^2} = 4{m^2} - 4$$

$$m = {2 \over {\sqrt 3 }}$$

$$C = {{ - 2} \over {\sqrt 3 }}$$

$$T:y = {2 \over {\sqrt 3 }}x - {2 \over {\sqrt 3 }}$$

$$P:{y^2} = 4x$$

$${y^2} = 4\left( {{{\sqrt 3 y + 2} \over 2}} \right)$$

$${y^2} - 2\sqrt 3 y - 4 = 0$$

Area

$${1 \over 2}\left| {\matrix{ 0 & 0 \cr {{x_1}} & {{y_1}} \cr {2\sqrt 2 } & 0 \cr {{x_2}} & {{y_2}} \cr 0 & 0 \cr } } \right|$$

$$ = \left| {{1 \over 2}\left( { - 2\sqrt 2 {y_1} + 2\sqrt 2 {y_2}} \right)} \right|$$

$$ = \sqrt 2 \left| {{y_2} - {y_1}} \right| = \sqrt 2 \sqrt {{{({y_1} + {y_2})}^2} - 4{y_1}{y_2}} $$

$$ = \sqrt {56} $$

$$ = 2\sqrt {14} $$