Perpendiculars are drawn from two points on the axis of the parabola $y^2=4ax$ to a tangent to the parabola. If the points are situated at the a distance $d$ from the focus and the lengths of the perpendiculars are $p_1$ and $p_2$ respectively, then $p_1^2- p_2^2 = ?$.
I have identified these points: Focus $\equiv (a,0) \\ A\equiv (a-d,p_1)\\B\equiv(a+d.-p_2) $
We know that the length of focal chord is $a(t_2-t_1)^2$ where $t_1$ and $t_2$ are parameters of the points. Since, $t_2 = \dfrac{-p_2}{2a}$ and $t_1= \dfrac{p_1}{2a}$, I equated $a(t_2-t_1)^2 $ distance formula between points A and B but still couldn't reach the anser.
How do I solve this question then?
Looks like you have misread the problem.
Let the tangent be $yt =x+at^2$.Let perpendiculars be drawn from $A(a-d,0)$ and $B(a+d,0)$ of lengths $p_1, p_2$ resp. Also, the foot of the perpendcicular $C$ from focus $S(a,0)$ coincides with the y-intercept of the tangent i.e.$(0,at^2)$. Finally let $D(-at^2,0)$ be the x-intercept of the tangent.
We have $SC = a\sqrt {1+t^2}$. From similarity of triangles we have
$\dfrac{p_1}{a(1+t^2)-d} = \dfrac{p_2}{a(1+t^2)+d} = \dfrac{a}{\sqrt{1+t^2}}= \lambda$
Now we easily get $\ p_2-p_1 = 2d \lambda$ and $\ p_1+p_2 = 2a \lambda (1+t^2)$
Hence $p_2^2-p_1^2 = 4ad \lambda^2 (1+t^2)=4a^3d$