If the normals at two points $P$ and $Q$ of a parabola $y^{2} = 4ax$ intersect at a third point $R$ on the curve, then the product of ordinates of $P$ and $Q$ is

Question

Dr. Abubacker · Accepted Answer

(d)
Let the coordinates of $P$ and $Q$ are $(at_{1}^{2},\ 2at_{1})$ and $(at_{2}^{2},\ 2at_{2})$ respectively. Then the coordinates of $R$ are $\{ 2a + a\left( t_{1}^{2} + t_{2}^{2} + t_{1}t_{2} ight),\  - at_{1}t_{2}(t_{1} + t_{2})\}$
Since, $R$ lies on the parabola
$	herefore a^{2}t_{1}^{2}t_{2}^{2}{(t_{1} + t_{2})}^{2} = 4a\lbrack 2a + a\{\left( t_{1} + t_{2} ight)^{2} - t_{1}t_{2}\}brack$
$\Rightarrow {(t_{1} + t_{2})}^{2}\left\{ t_{1}^{2}t_{2}^{2} - 4 ight\} + 4\left( t_{1}t_{2} - 2 ight) = 0$
$\Rightarrow t_{1}t_{2} = 2$
$\Rightarrow y_{1}y_{2} = \left( 2at_{1} ight)\left( 2at_{2} ight) = 4a^{2}t_{1}t_{2}$
$	herefore y_{1}y_{2} = 8a^{2}$

Q. If the normals at two points $P$ and $Q$ of a parabola $y^{2} = 4ax$ intersect at a third point $R$ on the curve, then the product of ordinates of $P$ and $Q$ is

Explanation

Reviewed by Dr. Abubacker

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