If $x = m y + c$ is a normal to the parabola $x^{2} = 4 a y$ , then the value of $c$ is

- 2 a m - a m^{3}

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2 a m + a m^{3}

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- \frac{2 a}{m} - \frac{a}{m^{3}}

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\frac{2 a}{m} + \frac{a}{m^{3}}

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Solution

The correct option is B $- 2 a m - a m^{3}$
Let point be $(2 a t, a t^{2})$
$\frac{d y}{d x} = \frac{2 x}{4 a} = \frac{2 (2 a t)}{4 a} = t$
$\Rightarrow$ slope of normal
$m = - \frac{1}{t}$
$t = - \frac{1}{m}$
$(y - a t^{2}) = - \frac{1}{t} (x - 2 a t)$
$y = a t^{2} - \frac{x}{t} + 2 a$
$\Rightarrow x = t y - a t^{3} - 2 a t$
$∴$ c is $- a m^{3} - 2 a m$

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Similar questions

If the equation of the normal is y = mx + c to the parabola $y^{2} = 4 a x$ , then find the value of 'c' in terms of a and m.

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}

y^{2} = 4 a x

△_{1}

(a m_{1}^{2}, 2 a m_{1}) (a m_{2}^{2}, 2 a m_{2}), (a m_{3}^{2}, 2 a m_{3})

△_{2}

(a m_{1} m_{2}, a (m_{1} + m_{2})) (a m_{2} m_{3}, a (m_{2} + m_{3}))

(a m_{3} m_{1}, a (m_{3} + m_{1}))

\frac{△_{1}}{△_{2}}

y^{2} = 4 a x

(a, 2 a)

x^{2} = 4 a y

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