The equation of the common normal to the parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ is

y = x

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x + y = a

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y = x - 3 a

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x + y = 3 a

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Solution

The correct option is D $x + y = 3 a$
Equation of common normal to $y^{2} = 4 a x$ and $x^{2} = 4 a y$ is :
Equation of normal to $y^{2} = 4 a x$ is
$y = m x - 2 a m - a m^{3} . . . . . . . . . (1) x^{2} = 4 a (m x - 2 a m - a m^{3}) x^{2} - 4 a m x + {8 a}^{2} m + 4 a^{2} m^{3} = 0$
Equal roots $D = 0$
$16 a^{2} m^{2} - 4 ({8 a}^{2} m + 4 a^{2} m^{3}) = 0 m^{2} - m + 2 = 0 m = 2, - 1$
Put $m = - 1$ in (1)
$y = - x + 2 a + a x + y = 3 a$

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