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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 $\triangle S P M$ is an equilateral triangle, then $P$ is equal to
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The correct answer is:
$(9,4 \sqrt{3})$
Given, that the $\triangle S P M$ (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, $P S=8$
$\sqrt{\left(3+2 t^2-5\right)^2+(4 t)^2}=8$
$\Rightarrow \quad \sqrt{\left(2 t^2-2\right)^2+(4 t)^2}=8$
$\Rightarrow \quad \sqrt{\left(2 t^2+2\right)^2}=8$
$\Rightarrow \quad 2 t^2+2=8$
$\Rightarrow \quad 2 t^2=6$
$\Rightarrow \quad t=\sqrt{3}$
$\therefore P(3+2 \times 3,4 \times \sqrt{3})=P(9,4 \sqrt{3})$
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, $P S=8$
$\sqrt{\left(3+2 t^2-5\right)^2+(4 t)^2}=8$
$\Rightarrow \quad \sqrt{\left(2 t^2-2\right)^2+(4 t)^2}=8$
$\Rightarrow \quad \sqrt{\left(2 t^2+2\right)^2}=8$
$\Rightarrow \quad 2 t^2+2=8$
$\Rightarrow \quad 2 t^2=6$
$\Rightarrow \quad t=\sqrt{3}$
$\therefore P(3+2 \times 3,4 \times \sqrt{3})=P(9,4 \sqrt{3})$
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