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