Question

The normal at the point P$(a{p}^2,2ap)$ meets the parabola $y^2=4ax$ again 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. $q^2=2$
• C. $p=2q$
• D. $q=2p$

Answer 1

Answer: (A)

$p^2=2$

Parabola: $y^2=4ax.....................\left(1\right)$

As we know if normal from $t_1$ cuts $t_2$ again at parabola, $t_2=-t_1-{\frac{2}{t}}_1$

So, $P(a{p}^2,2ap)$ and $Q(a{q}^2,2aq)=>q=-p-\frac{2}{p}$

Lines joining origin $O$ and $P$ and $Q$ are

$OP:y=\frac{2ap}{a{p}^2}x$

$=>py=2x.....................\left(3\right)$

Slope $m_1=\frac{2}{p}$

$OQ:y=\frac{2aq}{a{q}^2}x$

$y=\frac{2}{q}x.....................\left(4\right)$

$m_2=\frac{2}{q}$

As $\left(3\right)$ and $\left(4\right)$ are perpendicular to each other, $m_1{m}_2=-1$

$=>m_1{m}_2=\frac{2}{p}.\frac{2}{q}$

$=>-1=\frac{2}{p}.\frac{2}{(-p-\frac{2}{p})}$ (from $\left(2\right)$)

$=>p^2+2-4=0$

$=>p^2=2$