If a normal chord of a parabola $y^2=4 a x$ subtends a right angle at the origin, then the slope of that normal chord is

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