If two parabolas y^ 2 = 4a (x – k) and x^ 2 = 4a (y – k) have only...

Parabolas $y^{2} = 4a \left(x – k\right)$ and $x^{2} = 4a\left(y – k\right)$ touche each other at line $y = x$
($∵$ both parabola are inverse of each other)
$⇒y = x$ is a common tangent at $P$
$⇒$ we get point $P$ by solving $y^{2} = 4a \left(x – k\right)$ and y = x
⇒$point of contact$P$is \left(2a, 2a\right)<br />$⇒$equation of normal at$P \left(2a, 2a\right)$is$x + y = 4a