Question

If three normals are drawn from the point $(6,0)$ to the parabola $y^{2}=4 a x$ of which two are mutually perpendicular, then length of its latus rectum is

Answer 1

Normal of parabola $y^{2}=4 a x$ is
$y=m x-a m^{3}-2 a m$ and it is passing through $(6,0)$
$\Rightarrow 6 m-a m^{3}-2 a m=0 $
$\Rightarrow m=0,$ or, $ 6-2 a-a m^{2}=0$
$m=0 \Rightarrow $ normal is $x$ axis
$a m^{2}+2 a-6=0 $
$\Rightarrow \frac{2 a-6}{a}=-1 \left(\because m_{1} m_{2}=-1\right)$
We cannot draw a normal parallel to axis of parabola.
$2 a-6=0-a \Rightarrow 3 a=6 \Rightarrow a=2 $
$\Rightarrow $ Length of latus rectum $=8$