If normal at p to parabola $y^2=4ax$ meets parabola again at Q such that PQ subtends a right angle at the vertex of $y^2=4ax$ then p will be

The correct option is C

$\left(2\right(\sqrt{2}a,2a)$

It is useful to remember that fact that,If normal at

$(a{t}_1^2,2a{t}_1)$ meet the parabola at $(a{t}_2^2,2a{t}_2)$ then,

$t_2=-t_1-\frac{2}{t_1}$ .........(1)

Here, Given slope of PO=$\frac{2a{t}_1}{a{t}_1^2}=\frac{2}{t_1}$

Slope of $QO=\frac{2a{t}_2}{a{t}_2^2}=\frac{2}{t_2}$

And (slope of PO)(slope of QO)=-1

$\Rightarrow \left(\frac{2}{t_1}\right)\left(\frac{2}{t_2}\right)=-1\Rightarrow t_1{t}_2=-4$ ......(2)

From (1),$t_2=-t_1-\frac{2}{t_1}$

From (2),$t_2=-\frac{4}{t_1}$

$\Rightarrow \frac{4}{t_1}=t_1+\frac{2}{t_1}$

$4=t_1^2+2$

$\Rightarrow t_1=\pm \sqrt{2}$

$\Rightarrow p=(2at,a{t}^2)=\left(2a\right(\pm \sqrt{2}),a(2\left)\right)$

=$(\pm 2\sqrt{2}a,2a)$

$\therefore$ Only option c satisfies.