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

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

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

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Solution

The correct option is B
$y = m x - 2 a m - a m^{3}$

Given value of normal is m. we know at point (h,k) slope of tangent is $\frac{2 a}{k}$ . Therefore slope of normal would be $\frac{- k}{2 a}$ .

We have, $k^{2} = 4 a h a n d m = \frac{- k}{2 a}$

Substituting for c in y = mx + c at point (h,k)

k = m(h) + c

$- 2 a (m) = m (\frac{k^{2}}{4 a}) + c$

$- 2 a m - m (\frac{(- 2 a m)^{2}}{4 a}) = c$

we get $c = - 2 a m - a m^{3}$

Therefore equation of normal with slope m for the given parabola would be $y = m x - 2 a m - a m^{3}$

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