The normal $y = m x - 2 a m - a m^{3}$ to the parabola $y^{2} = 4 a x$ subtends a right angle at the vertex if

m = 1

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m = \sqrt{2}

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m = - \sqrt{2}

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m = \frac{1}{\sqrt{2}}

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Solution

The correct options are
B $m = - \sqrt{2}$
C $m = \sqrt{2}$
Making $y^{2} = 4 a x$ homogeneous with the help of
$y = m x - 2 a m - a m^{3}$
or $m x - y = 2 a m + a m^{3}$
or $\frac{m x - y}{2 a m + a m^{3}} = 1$
then $y^{2} = 4 a x \times (1)$
$y^{2} = 4 a x \times (\frac{m x - y}{2 a m + a m^{3}})$
$y^{2} = \frac{4 x (m x - y)}{(2 m + m^{2})}$
$\Rightarrow (2 m + m^{2}) y^{2} - 4 m x^{2} + 4 x y = 0$ ........ $(1)$
$∵$ angle between the lines represented by eqn $(1)$ is $\frac{π}{2}$ (given)
$∴$ coefficient of $x^{2} +$ coefficient of $x^{2} = 0$
$\Rightarrow - 4 m + (2 m + m^{3}) = 0$
$\Rightarrow m^{3} - 2 m = 0$
$\Rightarrow m (m^{2} - 2) = 0$
$\Rightarrow m \neq 0, m^{2} = 2$
$∴ m = \pm \sqrt{2}$

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