If the perpendicular 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.

Hint: In the above type of question first of all we will have to suppose any parabola for which the focus and its general point are already known and after that we will move further with the given condition.Also we will use the formula for the perpendicular distance from a given point which is given below;
\[\begin{align}
  & \text{let, }y=mx+c\text{ is the equation of line and (}{{x}_{1}},{{y}_{1}})\text{ is a point}\text{.} \\
 & \text{Then the perpendicular distance}=\dfrac{{{y}_{1}}-m{{x}_{1}}-c}{\sqrt{1+{{m}^{2}}}}. \\
\end{align}\]

Complete step by step answer:
Let us suppose a horizontal parabola whose equation is ${{y}^{2}}=4ax$ having focus (a, 0) and general point is \[(a{{t}^{2}},2at)\].
Let, the two fixed points on the axis be \[A(a-h,0)\text{ and }B(a+h,0)\]where h is any constant.
Also, the equation of tangent to parabola ${{y}^{2}}=4ax$ is $yt=x+a{{t}^{2}}$.
Let, \[{{p}_{1}}\text{ and }{{p}_{2}}\] be the perpendicular from point A and B respectively upon the tangent,
\[\begin{align}
  & \Rightarrow {{p}_{1}}=\dfrac{a+h+a{{t}^{2}}}{\sqrt{1+{{t}^{2}}}} \\
 & \Rightarrow {{p}_{2}}=\dfrac{a-h+a{{t}^{2}}}{\sqrt{1+{{t}^{2}}}} \\
\end{align}\]
Now, we will find the difference of the square of the distance which is shown below;
\[\begin{align}
  & {{p}_{1}}^{2}-{{p}_{2}}^{2}={{\left( \dfrac{a+h+a{{t}^{2}}}{\sqrt{1+{{t}^{2}}}} \right)}^{2}}-{{\left( \dfrac{a-h+a{{t}^{2}}}{\sqrt{1+{{t}^{2}}}} \right)}^{2}} \\
 & \Rightarrow {{p}_{1}}^{2}-{{p}_{2}}^{2}=\dfrac{{{a}^{2}}+{{h}^{2}}+{{a}^{2}}{{t}^{4}}+2ah+2ah{{t}^{2}}+2{{a}^{2}}{{t}^{2}}-{{a}^{2}}-{{h}^{2}}-a{{t}^{4}}+2ah+2ah{{t}^{2}}-2{{a}^{2}}{{t}^{2}}}{1+{{t}^{2}}} \\
 & \Rightarrow {{p}_{1}}^{2}-{{p}_{2}}^{2}=\dfrac{4ah+4ah{{t}^{2}}}{1+{{t}^{2}}}=\dfrac{4ah(1+{{t}^{2}})}{1+{{t}^{2}}} \\
 & \Rightarrow {{p}_{1}}^{2}-{{p}_{2}}^{2}=4ah \\
\end{align}\]
Here, we get the difference of the square of perpendicular distance is equal to \[4ah\] which is a constant.
 Here, it is proved that the difference of the square of the perpendicular distance from the two fixed points on the axis of parabola is constant.
NOTE: Remember the perpendicular distance formula as well as the equation of the parabola and properties related with it as it makes the question easy to understand.