Question

If perpendiculars be drawn on any tangent to a parabola from two fixed points on the axis, which are equidistant from the focus, prove that the difference of their squares is constant.

Answer 1

Focus of the parabola is $S(a,0)$

Let the two fixed point on the axis be $A(a+h,0)$ and $B(a-h,0)$

Equation of tangent to parabola $y^2=4ax$ is $ty=x+a{t}^2$

Let $p_1$ and $p_2$ be the perpendicular from $A$ and $B$ respectively upon the tangent

$p_1=\frac{a+h+a{t}^2}{\sqrt{1+t^2}}{p}_2=\frac{a-h+a{t}^2}{\sqrt{1+t^2}}{p}_1^2-p_2^2={\left(\frac{a+h+a{t}^2}{\sqrt{1+t^2}}\right)}^2-{\left(\frac{a-h+a{t}^2}{\sqrt{1+t^2}}\right)}^2{p}_1^2-p_2^2=\frac{a^2+h^2+a^2{t}^4+2ah+2ah{t}^2+2a^2{t}^2-a^2-h^2-a^2{t}^4+2ah+2ah{t}^2-2a^2{t}^2}{1+t^2}{p}_1^2-p_2^2=\frac{4ah+4ah{t}^2}{1+t^2}=\frac{4ah(1+t^2)}{1+t^2}{p}_1^2-p_2^2=4ah$

As $h$ is fixed ${\therefore p}_1^2-p_2^2$ is constant

Hence proved