First slide

Parabola

Question

If the normal at $P$ to the parabola $y^{2} = 4 a x$ meets
the parabola again at $Q$ such that $P Q$ subtends a
right angle at the vertex of the parabola, then the
coordinates of $P$ are

Moderate

A

$(2 a, 2 \sqrt{2} a)$
B

$(- 2 a, 2 \sqrt{2} a)$
C

$(2 \sqrt{2} a, 2 a)$
D

$(2 \sqrt{2} a, - 2 a)$

Solution

Let the coordinates of $P$ be $(a m^{2}, - 2 a m)$ and $Q$ be
$\begin{array}{l} (a m^{2}, - 2 a m^{'}) then \frac{- 2 a m}{a m^{2}} \times \frac{- 2 a m^{'}}{a m^{' 2}} = - 1 \\ \Rightarrow m m^{'} = - 4 \end{array}$
Normal at $P, y = m x - 2 a m - a m^{3}$ Since it passes
through $Q$
$- 2 a (- \frac{4}{m}) ⊨ a m (\frac{16}{m^{2}}) - 2 a m - a m^{3} = 0$
$\Rightarrow m^{4} + 2 m^{2} - 8 = 0 \Rightarrow m = \pm \sqrt{2}$ and the
coordinates of $P$ are $(2 a, \pm 2 \sqrt{2} a)$ .

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