A normal is drawn at the point P to the parabola y^2=8 x , which is...

For the parabola $y^2=4 a x$, the equation of normal at $P$\left(a m²,-2 a m\right)$is$y=m x-2 a m-a m³$. Here,$m=\tan 60^{{}^{\circ}}=\sqrt{3}\thereforeP \equiv\lleft(a(\sqrt{3})^2,-2 $a(\sqrt{3})) \equiv(6,-4 \sqrt{3}) \quad(\becausea=$2)$Thus,$P$satisfies$2 x-y-12-4 $\sqrt{3}=$0