Question

The normal at the point $P(ap$${}^2$, $2ap)$ meets the parabola $y^2=4ax$ again at $Q(aq$${}^2$, $2aq)$ such that the lines joining the origin to $P$ and $Q$ are at right angle. Then

• A. $p^2=2$
• B. $q^2=2$
• C. $p=2q$
• D. $q=2p$

Answer 1

Answer: (A)

$p^2=2$

Given the equation of parabola is
$y^2=4ax$

Equation of normal at $P(a{p}^2,2ap)$ is given by
$y=-px+2ap+a{p}^3$

Since, it meet the parabola again at $Q(a{q}^2,2aq)$ is given by
$q=-p-\frac{2}{p}$

.....(i)
Slope of OP $=\frac{2ap-0}{a{p}^2-0}=\frac{2}{p}$

Slope of OQ $=\frac{2aq-0}{a{q}^2-0}=\frac{2}{q}$

Since OP $\perp$ OQ.

$\Rightarrow m_1{m}_2=-1$

$\Rightarrow pq=-4$

$\Rightarrow p(-p-\frac{2}{p})=-4$

$\Rightarrow p^2=2$