Let $M$ be the foot of the perpendicular from a point $P$ on the parabola $y^2=8(x-3)$ onto its directrix and let $S$ be the focus of the parabola. If $ riangle S P M$ is an equilateral triangle, then $P$ is equal to

Question

Let $M$ be the foot of the perpendicular from a point $P$ on the parabola $y^2=8(x-3)$ onto its directrix and let $S$ be the focus of the parabola. If $	riangle S P M$ is an equilateral triangle, then $P$ is equal to

Aicharya · Accepted Answer

Given, that the $	riangle S P M$ (which is shown in figure) is equilateral. Also, given parabola is $y^2=8(x-3)$ focus of this parabola is $S(5,0)$ and vertex $A(3,0)$. Let coordinate of $P\left(h+a t^2, k+2 a tight)$ $=P\left(3+2 t^2, 4 tight)$ Then, coordinate of $M(-5,4 t)$. We know that the side of this equilateral triangle is $4 a=4 	imes 2=8$ Now, $P S=8$ $\sqrt{\left(3+2 t^2-5ight)^2+(4 t)^2}=8$ $\Rightarrow \quad \sqrt{\left(2 t^2-2ight)^2+(4 t)^2}=8$ $\Rightarrow \quad \sqrt{\left(2 t^2+2ight)^2}=8$ $\Rightarrow \quad 2 t^2+2=8$ $\Rightarrow \quad 2 t^2=6$ $\Rightarrow \quad t=\sqrt{3}$ $	herefore P(3+2 	imes 3,4 	imes \sqrt{3})=P(9,4 \sqrt{3})$

Let $M$ be the foot of the perpendicular from a point $P$ on the parabola $y^2=8(x-3)$ onto its directrix…

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