The equation of normal to the curve $y^{2} = 16 x$ at the point $(1, 4)$ is

2 x + y = 6

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2 x - y + 2 = 0

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x + 2 y = 9

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None of these

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Solution

The correct option is C $x + 2 y = 9$
$y^{2} = 16 x$
$\Rightarrow 2 y \frac{d y}{d x} = 16$
$\Rightarrow∴ \frac{d y}{d x} = \frac{8}{y}$
Thus slope of tangent at $(1, 4)$ is $m = 2$
$∴$ Slope is normal is $= - \frac{1}{2}$
Therefore, the equation of normal at $(1, 4)$ is given by, $(y - 4) = - \frac{1}{2} (x - 1)$
$\Rightarrow x + 2 y = 9$
Hence, option 'C' is correct.

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