If $(V)$ and $a = 1$ is satisfied, then the equation of $S^{'} = 0$ is

y^{2} = 4 μ x

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x^{2} = 4 μ y

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y^{2} = 2 μ x

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x^{2} = 2 μ y

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Solution

The correct option is A $x^{2} = 2 μ y$
The equation of the normal to $y^{2} = 4 a x$ is $y = m x - 2 a m - a m^{3}$ .......... $(1)$
$∵$ it passes through $(h, k)$ then $a m^{3} + m (2 a - h) + k = 0$ ........... $(2)$
$∵$ Roots of eqn $(2)$ be
$m_{1}, m_{2}, m_{3}$ then
$m_{1} + m_{2} + m_{3} = 0$ ............ $(3)$
$m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{(2 a - h)}{a}$ ............. $(4)$
and $m_{1} m_{2} m_{3} = - \frac{k}{a}$ ..... $(5)$
The equation of the line $A B$ is
$y - (- 2 a m_{1}) = \frac{(- 2 a m_{2}) - (- 2 a m_{1})}{a m_{2}^{2} - a m_{1}^{2}} (x - a m_{1}^{2})$
$\Rightarrow y + (- 2 a m_{1}) = - \frac{2}{m_{1} + m_{2}} (x - a m_{1}^{2})$
$\Rightarrow y (m_{1} + m_{2}) + 2 a m_{1} (m_{1} + m_{2}) = - 2 x + 2 a m_{1}^{2}$
$\Rightarrow y (m_{1} + m_{2}) + 2 a m_{1} m_{2} = - 2 x$
From eqns $(3)$ and $(5)$ replacing $k$ by $μ$ (because $y = μ = k$ )
$\Rightarrow y (- m_{3}) + 2 a (- \frac{μ}{a m_{3}}) = - 2 x$
$\Rightarrow y m_{3}^{2} - 2 m_{3} x + 2 μ = 0$ which is a quadratic in $m_{3}$
Since, $A B$ will touch it then
$B^{2} - 4 A C = 0$
$\Rightarrow {(- 2 x)}^{2} - 4 y .2 μ = 0$
or $x^{2} = 2 μ y$

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