The locus of the orthocenter of the triangle formed by the focal chord of the parabola $y^{2} = 4 a x$ and the normal drawn at its extermities is

$y^{2} = a (x - 3 a)$

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$y^{2} = a (x + 3 a)$

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$y^{2} = a (x - 4 a)$

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$y^{2} = a (x + 4 a)$

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Solution

The correct option is A
$y^{2} = a (x - 3 a)$

The normals at the extremities of focal chord meet at right angle. So orthocenter is the point of intersection of normals.

If $P (a t_{1}^{2}, 2 a t_{1}), Q (a t_{2}^{2}, 2 a t_{2})$ then $t_{1} t_{2} = - 1$ point of intersection of normals.

$h = a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2} + 2)$

$k = - a t_{1} t_{2} (t_{1} + t_{2})$

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Similar questions

The equation of the locus of the point of intersection of two normals to the parabola $y^{2} = 4 a x$ which are perpendicular to each other is

y^{2} = 4 a x

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

y^{2} = 4 a x

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