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 $\Delta S P M$ is an equilateral triangle, then $P$ is equal to

• A. $(4 \sqrt{3}, 8)$
• B. $(8,4 \sqrt{3})$
• C. $(9,4 \sqrt{3})$
• D. $(4 \sqrt{3}, 9)$

Answer 1

Answer: (C)

Given, that the $\Delta S P M$ (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 $P\left(h +a t^{2}, k+2 a t\right)$
$=P\left(3+2 t^{2}, 4 t\right)$
Then, coordinate of $M(-5,4 t)$.
We know that the side of this equilateral triangle is $4 a=4 \times 2=8$
Now, $P S=8$
$\sqrt{\left(3+2 t^{2}-5\right)^{2}+(4 t)^{2}}=8$
$\Rightarrow \sqrt{\left(2 t^{2}-2\right)^{2}+(4 t)^{2}}=8$
$\Rightarrow \sqrt{\left(2 t^{2}+2\right)^{2}}=8$
$\Rightarrow 2 t^{2}+2=8$
$\Rightarrow 2 t^{2}=6$
$\Rightarrow t=\sqrt{3}$
$\therefore P(3+2 \times 3,4 \times \sqrt{3})=P(9,4 \sqrt{3})$