A normal is drawn at the point $P$ to the parabola $y^2=8 x$, which is inclined at $60^{\circ}$ with the…
A normal is drawn at the point $P$ to the parabola $y^2=8 x$, which is inclined at $60^{\circ}$ with the straight line $y=8$. Then the point $P$ lies on the straight line
$2 x+y-12-4 \sqrt{3}=0$
$2 x-y-12+4 \sqrt{3}=0$
$2 x-y-12-4 \sqrt{3}=0$
None of these
Solution
For the parabola $y^2=4 a x$, the equation of normal at $P\left(a m^2,-2 a m\right)$ is $y=m x-2 a m-a m^3$.
Here, $m=\tan 60^{\circ}=\sqrt{3}$
$\therefore P \equiv\left(a(\sqrt{3})^2,-2 a(\sqrt{3})\right) \equiv(6,-4 \sqrt{3}) \quad(\because a=2)$
Thus, $P$ satisfies $2 x-y-12-4 \sqrt{3}=0$