Question

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

• A. $(4a,4a)$
• B. $(3a,2\sqrt{3}a)$
• C. $(\frac{a}{3},\frac{2a}{\sqrt{3}})$
• D. $(3a,-2\sqrt{3}a)$

Answer 1

Answer: (B), (C)

The correct options are
B $(3a,2\sqrt{3}a)$
C $(\frac{a}{3},\frac{2a}{\sqrt{3}})$

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