Normal at a point $P$on the parabola $y^2=4ax$ meets
the axis at $Q$such that the distance of $Q$ from the
focus of the parabola is 10$a$. The coordinates of $P$
are:

Equation of the normal at $P\left(a{t}^2,2at\right)$ to the parabola

$y^2=4ax\text{ is }y=-tx+2at+a{t}^3$ which meets the

axis $y=0$ at $x=a\left(2+t^2\right)$. So coordinates of $Q$are

$\left(a\left(2+t^2\right),0\right).$Focus is $S(a,0)$

$QS=2a+a{t}^2-a=a\left(1+k^2\right)=10a\Rightarrow t^2=9$

So the required coordinates are $(9a,\pm 6a)$.