Find the equation of all normal to the parabola $y^{2} = 4 a x$ which pass through the point $(3 a, 0)$

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Solution

Equation of the normal is given by,

$y - y_{1} = - \frac{y_{1}}{2 a} (x - x_{1})$

Given, $(x_{1}, y_{1}) = (3 a, 0)$

$\Rightarrow y - 0 = - \frac{0}{2 a} (x - 3 a)$

$∴ y = 0$ is only the normal equation to the parabola

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