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Q.
Two lines are drawn from point $P (\alpha, \beta)$ which touches $y ^2=8 x$ at $A , B$ and touches $\frac{x^2}{4}+\frac{y^2}{6}=1$ at $C, D$, then
Conic Sections
Solution:
$\frac{x^2}{4}+\frac{y^2}{6}=1$ and $y^2=8 x$
Tangent to the ellipse
$y=m x \pm \sqrt{4 m^2+6}$
Tangent to the parabola
$\text { Tangent to the parabola } $
$y = mx +\frac{2}{ m } $
$\therefore \frac{2}{ m }= \pm \sqrt{4 m ^2+6} $
$4= m ^2\left(4 m ^2+6\right) $
$2 m ^4+3 m ^2-2=0 $
$\left( m ^2+2\right)\left(2 m ^2-1\right)=0 \Rightarrow m = \pm \frac{1}{\sqrt{2}} $
$\therefore y =\frac{ x }{\sqrt{2}}+2 \sqrt{2} $
$\sqrt{2} y = x +4 \Rightarrow \alpha=-4, \beta=0 $
$\operatorname{Ar}(\Delta PAB )=\frac{ s _1^{3 / 2}}{2| a |}=\frac{(0+32)^{3 / 2}}{2 \cdot 2}=\frac{32 \cdot 4 \sqrt{2}}{4}=32 \sqrt{2}$
