The normal at three points $P, Q, R$ of the parabola $y^{2} = 4 a x$ meet in $(h, k)$ . The centroid of $△ P Q R$ lies on

y = 0

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x = 0

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x = - a

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x = a

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Solution

The correct option is B $y = 0$
The equation of any normal to the parabola $y^{2} = 4 a x$ is $y = m x - 2 a m - a m^{3}$
If it passes through $(h, k)$ then $k == m h - 2 a m - a m^{3}$
$\Rightarrow a m^{3} + (2 a - h) m + k = 0$
This is a cubic in $m$ . Let its roots be $m_{1}, m_{2}, m_{3}$
Then, $m_{1} + m_{2} + m_{3} = 0; m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a}$ ...(1)
Co-ordinates of $P, Q, R$ , the feet of the three normals are
$(a {m_{1}}^{2} - 2 a m_{1}); (a {m_{2}}^{2} - 2 a m_{2}); (a {m_{3}}^{2} - 2 a m_{3})$ respectively
Let $G \equiv (¯ ¯ ¯ x, ¯ ¯ ¯ y)$ be the centroid of $△ P Q R$
Then $¯ ¯ ¯ x = \frac{a {m_{1}}^{2} + a {m_{2}}^{2} + a {m_{3}}^{2}}{3} = \frac{a}{3} ({m_{1}}^{2} + {m_{2}}^{2} + {m_{3}}^{2})$
$= \frac{a}{3} [{(0)}^{2} - 2 (\frac{2 a - h}{a})] = \frac{2}{3} (h - 2 a)$
$¯ ¯ ¯ y = \frac{(- 2 a m_{1}) + (- 2 a m_{2}) + (- 2 a m_{3})}{3} = - \frac{2 a}{3} (m_{1} + m_{2} + m_{3}) = 0$
$\Rightarrow$ centroid of $△ P Q R$ is $G \equiv [\frac{2}{3} (h - 2 a), 0]$ which lies on $y = 0$

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