A point on a parabola, the foot of the perpendicular from it upon the directrix, and the focus are the vertices of an equilateral triangle. Prove that the focal distance of the point is equal to the latus rectum.

Hint: Look at the figure and if from point P a perpendicular is drawn to the directrix, then the foot of perpendicular will be at M. Now, analyze the conditions and apply distance formula to find the solution.

Complete step by step answer:

For parabola ${y^2} = 4ax$

Let from any point P $({a{t^2},2at})$ a perpendicular is drawn to the directrix.

Then you can see that foot of perpendicular will be

M (- a,2at)

Focus is at S (a,0)

As $\Delta PMS$ is equilateral.

Therefore, MS = SP

$\Rightarrow M{S^2} = S{P^2}$

Now using Distance formula we have,

${\left( {\sqrt {{{\left( { - a - a} \right)}^2} + {{\left( {2at - 0} \right)}^2}} } \right)^2} = {\left( {\sqrt {{{\left( {a - a{t^2}} \right)}^2} + {{\left( {0 - 2at} \right)}^2}} } \right)^2}$

$\Rightarrow {\left( { - a - a} \right)^2} + {\left( {2at - 0} \right)^2} = {\left( {a - a{t^2}} \right)^2} + {\left( {0 - 2at} \right)^2}$

$\Rightarrow 4{a^2} + 4{a^2}{t^2} = {\left( {a - a{t^2}} \right)^2} + 4{a^2}{t^2}$

$\Rightarrow 4{a^2} = {\left( {a - a{t^2}} \right)^2}$

$\Rightarrow a - a{t^2} = \pm 2a$

$\Rightarrow - a{t^2} = a, - 3a$

$\Rightarrow {t^2} = - 1,3$

As square of something cannot be negative, hence

${t^2} = 3{\text{ and }}{t^2} \ne - 1$

Focal distance of $P = a + a{t^2} = a + 3a = 4a$

And length of a latus rectum is always = 4a

Hence Proved.

Note: This is a general question and this way of doing it must be remembered. It is advised that the student remembers that focal distance of any point on parabola is equal to the length of the latus rectum as it is used in many questions.