Question

If the normal at the point $P(a{p}^2,2ap)$ meets the parabola at $Q(a{q}^2,2aq)$ such that the lines joining the origin to $P$ and $Q$ are at right angle, then

• A. $p^2=2$
• B. $pq-4=0$
• C. $pq+4=0$
• D. $p^2=4$

Answer 1

Answer: (A), (C)

$pq+4=0$

Equation of the normal at $(a{p}^2,2ap)$ to $y^2=4ax$ is
$y+px=2ap+a{p}^3$
It passes through $Q(a{q}^2,2aq)$, so
$\therefore 2aq+a{q}^{2}p=2ap+a{p}^3\Rightarrow (p-q)(2+p^2+pq)=0[\because a\ne 0]\therefore 2+p^2+pq=0\cdots \left(1\right)[\because p\ne q]$

Now $OP\perp OQ$, so
$\therefore \frac{2}{p}\times \frac{2}{q}=-1\Rightarrow pq=-4\Rightarrow p^2=2\left(\text{using}1\right)$