The equation of the normal at the ends of the latusrectum of the parabola $y^{2} = 4 a x$ by

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

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x^{2} + y^{2} - 6 a y + {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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none of these

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Solution

The correct option is A $x^{2} - y^{2} - 6 a x + {9 a}^{2} = 0$
$y^{2} = a x$ $
$2 y \frac{d y}{d x} = 4 a$
$\frac{d y}{d x} = \frac{2 a}{y}$
slope of $(a_{1} - 2 a)$
Slope = $\frac{- 2 a}{2 a} = - 1$
slope of normal $= 1$
$\frac{y + 2 a}{x - a} = 1$
for $(a, 2 a)$
$\frac{d y}{d x} = \frac{2 a}{2 a} = 1$
slope of normal $= - 1$
$\frac{y - 2 a}{x - a} = - 1$
$y - 2 a = - x + 9$
$x + y - 3 a = 0$
pair of straight lin.
$(x - y - 3 a) (x + y - 3 a) = 0$
$x^{2} - y^{2} - 6 a n + 9 a^{2} = 0$

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