The equation of the normals at the ends of the latus rectum of the parabola $y^{2} = 4 a x$ are given by

x^{2} - y^{2} - 6 a x + 9 a^{2} = 0

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x^{2} - y^{2} - 6 a x - 6 a y + 9 a^{2} = 0

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x^{2} - y^{2} - 6 a y + 9 a^{2} = 0

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Solution

The correct option is C $x^{2} - y^{2} - 6 a x + 9 a^{2} = 0$
The coordinate of the ends of the latus rectum of the parabola $y^{2} = 4 a x$ are $(a, 2 a)$ and $(a, - 2 a)$ respectively.
The equation of the normal at $(a, 2 a)$ to $y^{2} = 4 a x$ is
$y - y_{1} = - \frac{y_{1}}{2 a} (x - x_{1}) \Rightarrow y - 2 a = - \frac{2 a}{2 a} (x - a)$
$\Rightarrow x + y - 3 a = 0$ ...(1)
Similarly the equation of the normal $(a, - 2 a)$ is $x - y - 3 a = 0$ ...(2)
The combined equation of (1) and (2) is $x^{2} - y^{2} - 6 a x + 9 a^{2} = 0$

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