If a normal chord at a point $t$ on the parabola $y^2=4 a x$ subtends a right angle at the vertex, then $t$ equals to

The perpendicular of the normal to the parabola $y^2=4 a x$ at $P$ is
Suppose, it meets the parabola at $Q$. If $O$ be the vertex of the parabola, then the combined equation of $O P$ and $O Q$ is a homogeneous equation of second degree. $$ \begin{aligned} & \quad y^2=4 a x\left(\frac{y+t x}{2 a t+a t^3}\right) \\ & \Rightarrow y^2\left(2 a t+a t^3\right)=4 a x(y+t x) \\ & \Rightarrow 4 a t x^2+4 a x y-\left(2 a t+a t^3\right) y^2=0 \end{aligned} $$ Since, $O P$ and $O Q$ are at right angles, then Coefficient of $x^2+$ Coefficient of $y^2=0$ $$ \begin{aligned} \therefore & 4 a t-2 a t-a t^3 & =0 \\ \Rightarrow & t^2=2 \Rightarrow t & =\sqrt{2} \end{aligned} $$