Consider $P$ is a point on $y^{2} = 4 a x,$ if the normal at $P,$ the axis and the focal radius of $P$ form an equilateral triangle. Then coordinates of $P$ are

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Solution

Let $P$ be $(a t^{2}, 2 a t)$ and $P N$ be normal
$\frac{\sqrt{3}}{2} S P = P M$ , where $M$ is foot of perpendicular from $P$ upon axis of parabola
$\Rightarrow \frac{\sqrt{3}}{2} (a t^{2} + a) = 2 a t$
$\Rightarrow \sqrt{3} t^{2} + \sqrt{3} = 4 t$
$\Rightarrow \sqrt{3} t^{2} - 4 t + \sqrt{3} = 0$
$\Rightarrow \sqrt{3} t^{2} - 3 t - t + \sqrt{3}$
$\Rightarrow \sqrt{3} t (t - \sqrt{3}) - 1 (t - \sqrt{3}) = 0$
$\Rightarrow (t - \sqrt{3}) (\sqrt{3} t - 1) = 0$
$∴ t = \sqrt{3}$ or $t = \frac{1}{\sqrt{3}}$
So, $P$ is $(3 a, 2 \sqrt{3} a) or (\frac{a}{3}, \frac{2 a}{\sqrt{3}})$

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