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.

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Solution

For $y^{2} = 4 a x$
Let from any point $P (a t^{2}, 2 a t)$ perpendicular is drawn to the directrix
Then you can clearly see from the the figure feet of perpendicular will be $M (- a, 2 a t)$
Focus is $S (a, 0)$
As $△ P M S$ is equilateral $∴ M S = M P$
${M S}^{2} = {S P}^{2} {(- a - a)}^{2} + {(2 a t - 0)}^{2} = {(a - a t^{2})}^{2} + {(0 - 2 a t)}^{2} 4 a^{2} + 4 a^{2} t^{2} = {(a - a t^{2})}^{2} + 4 a^{2} t^{2} {(a - a t^{2})}^{2} = 4 a^{2} a - a t^{2} = \pm 2 a - a t^{2} = a, - 3 a t^{2} = - 1, 3$
square of something equal to negative is not possible
$\Rightarrow t^{2} = 3$
Focal distance of $P = a + a t^{2} = a + 3 a = 4 a$
Laturectum $= 4 a$
Hence proved.

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