Normal to the parabola $y^{2} = 4 a x$ where $m$ is the slope of the normal is

y = m x + 2 a m - a m^{3}

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y = m x - 2 a m - a m^{3}

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y = m x - 2 a m + a m^{3}

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Solution

The correct option is D $y = m x - 2 a m - a m^{3}$
Given, $y^{2} = 4 a x$
$\Rightarrow 2 y d y = 4 a$
$\Rightarrow \frac{d y}{d x} = \frac{2 a}{y}$
Thus slope of normal to the given parabola at any point is,
$= - ⎛ ⎜ ⎝ \frac{1}{\frac{d y}{d x}} ⎞ ⎟ ⎠ = - \frac{y}{2 a} = m$ (given)
$\Rightarrow y = - 2 a m$
Thus from $y^{2} = 4 a x,$ $x = \frac{y^{2}}{4 a} = a m^{2}$
Therefore the point is $(a m^{2}, - 2 a m)$
Hence the required line passing through $(a m^{2}, - 2 a m)$ and having slope $m$ is,
$(y + 2 a m) = m (x - a m^{2})$
$\Rightarrow y = m x - 2 a m - a m^{3}$

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Similar questions

Given slope m, find the equation of normal having slope m to the parabola $y^{2} = 4 a x$

y = m x - 2 a m - a m^{3}

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