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.

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Solution

The correct option is C

Let tha equation of the normal be $y = m x + c$

Equation of the tangent on the parabola $y^{2} = 4 a x$ at $P (x_{1}, y_{1})$

$y y_{1} = 2 a (x + x_{1})$

$y = \frac{2 a}{y_{1}} x + \frac{2 a}{y_{1}} x_{1}$

Slope of tangent $= \frac{2 a}{y_{1}}$

Slope of the normal $= \frac{- 1}{m_{1}} = \frac{- 1}{\frac{2 a}{y_{1}}} = \frac{- y_{1}}{2 a}$

Slope of the normal $m = \frac{- y_{1}}{2 a}$

$\Rightarrow y_{1} = - 2 a m$

$y^{2} = 4 a x$

$P (x_{1}, y_{1})$ should satisfy the parabola

$y_{1}^{2} = 4 a x_{1}$

Substituting $y_{1} = - 2 a m$

We get,

$(- 2 a m)^{2} = 4 a x_{1}$

$4 a^{2} m^{2} = 4 a x_{1}$

$x_{1} = a m^{2}$

Coordinates of point $P (a m^{2}, - 2 a m)$ where m is the slope of the normal

The equation of normal $y = m x + c$

point P should satisfy the equation of the normal

$- 2 a m = m (a m^{2}) + c$

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

Suggest Corrections

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