If the line x-y+k=0 is a normal to $y^{2} = 4 a x$ then the value of k is

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-a

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-5a

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-3a

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Solution

The correct option is D
-3a

x-y+k=0 $y^{2} = 4 a x$
y=x+k
y=mx+c
$c = - 2 a m - a m^{3}$
$= - 2 a .1 - a {.1}^{3}$
C=-3a

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