The normals at point A and B on $y^{2} = 4 a x$ meet the parabola in a point C on the parabola, the locus of orthocenter of triangle ABC is the parabola

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$y^{2} = a (x + 6 a)$

$y^{2} = (x - 6 a)$

$y^{2} = a (x + 6)$

$y^{2} = (x + 6 a)$

answer is A.

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Detailed Solution

$y^{2} = 4 a x . . . . . . (1)$
$Let assume A and B are on the parabola$
$A (a t_{1}^{2}, 2 a t_{1}) B (a t_{2}^{2}, 2 a t_{2})$
$Equation of Normal at A .$
$t_{1} x + y = 2 a t_{1} + a t_{1}^{3} .....................(2)$
$Equation of normat at B,$
$t_{2} x + y = 2 a t_{2} + a {t^{3}}_{2}$
$Now equp (2) and equ (3) meers at C on parabola, t_{1} t_{2} = 2$
$slope of line ⊥ to the normal drawn at pt A = \frac{1}{t_{1}}$
$and this ⊥ line passes through pt B$
$\begin{array}{l} y - 2 a t_{2} = \frac{1}{t_{1}} (x - a t_{2}^{2}) \\ \Rightarrow t_{1} y - 2 a t_{1} t_{2} = x - a t_{2}^{2} \\ \Rightarrow x - t_{1} y = a t_{2}^{2} - 4 a [t_{1} t_{2} = 2] \end{array}$
$\begin{array}{l} y - 2 a_{1} = \frac{1}{t_{2}} (x - a t_{1}^{2}) \\ \Rightarrow t_{2} y - 2 a t_{1} t_{2} = x - a t_{1}^{2} \\ \Rightarrow x - t_{2} y = a t_{1}^{2} - 4 a (t_{1} t_{2} = 2) \end{array}$
$For orthocentre O (h, k), solve equ A D and B E$
$\begin{array}{l} x = a t_{1} (t_{1} + t_{2}) + a t_{2}^{2} - 4 a \\ x = a t_{1}^{2} + a t_{1} t_{2} + a t_{2}^{2} - 4 a \\ x = a t_{1}^{2} + a t_{2}^{2} - 2 a \\ x = a (t_{1}^{2} + t_{2}^{2}) - 2 a \\ x + 2 a = a (t^{2} + t_{2}^{2}) \\ (h_{,} k) = [a ({t_{1}}^{2} + t_{2}^{2}) - 2 a, a (t_{1} + t_{2})] \end{array}$
$\begin{array}{l} \Rightarrow h = a (t_{1}^{2} + t_{2}^{2}) - 2 a \\ h + 2 a = a (t_{1}^{2} + t_{2}^{2}) = a [t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2} - 2 t_{1} t_{2}] \\ h + 2 a = a [{(t_{1} + t_{2})}^{2} - 2 \times 2] \\ k = a (t_{1} + t_{2}) \\ \Rightarrow h + 2 a = a [\frac{k^{2}}{a^{2}} - 4] \\ t_{1} + t_{2} = \frac{k}{a} \\ \Rightarrow h + 2 a = a [\frac{k^{2} - 4 a^{2}}{a^{2}}] = \frac{k^{2} - 4 a^{2}}{a} \\ \Rightarrow a h + 2 a^{2} = k^{2} - 4 a^{2} \end{array}$
$y^{2} = a [x + 6 a]$