The equation of the normal to the curve $y^{2} = 4 a x$ at point $(a, 2 a)$ is

x - y + a = 0

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x + y - 3 a = 0

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x + 2 y + 4 a = 0

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x + y + 4 a = 0

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Solution

The correct option is C $x + y - 3 a = 0$
$y^{2} = 4 a x$
$2 y \frac{d y}{d x} = 4 a \Rightarrow \frac{d y}{d x} = \frac{2 a}{y}$
$∴ {(\frac{d y}{d x})}_{(a, 2 a)} = {(\frac{2 a}{y})}_{(a, 2 a)} = 1 = m$ (say)
$∴$ Slope of normal at the given point is $= - \frac{1}{m} = - 1$
Therefore, the equation of normal is, $(y - 2 a) = - 1 (x - a)$
$\Rightarrow x + y = 3 a$
Hence, option 'B' is correct.

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