Let E denote the parabola y 2 = 8 x . Let P = - 2 , 4 , and let Q and Q ' be two distinct points on E such…

$y^2=8x$

$\Rightarrow a=2,$ Focus $F\left(2,0\right),$ and Directrix $x=-2$

$\because P\left(-2,4\right)$ lies on directrix

$\Rightarrow Q{Q}^{\text{'}}$ will be a focal chord and tangents are mutually perpendicular.

$\Rightarrow \Delta QP{Q}^{\text{'}}$ is right angled.

Also the portion between point of contact and directrix of parabola subtends right angle at the focus.

So, $△PFQ$ is a right- angled triangle.

$PF=\sqrt{{\left(-2-2\right)}^2+{\left(4-0\right)}^2}=4\sqrt{2}$