If the tangents at two points of a parabola $y^{2} = 4 a x$ intersect at $(h, k),$ then find the co-ordinates of intersection point of normals at those two points.

(2 a - 2 h + \frac{k^{2}}{a}, \frac{h k}{a})

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(2 a - h + \frac{k^{2}}{a}, - \frac{h k}{a})

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(a + \frac{h}{2} + 2 \frac{k^{2}}{a}, - 2 \frac{h k}{a})

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(- a + 3 \frac{h}{2} + 3 \frac{k^{2}}{2 a}, - \frac{h k}{a})

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Solution

The correct option is A $(2 a - h + \frac{k^{2}}{a}, - \frac{h k}{a})$
For tangents $h = a t_{1} t_{2}, k = a (t_{1} + t_{2})$
For normals $x = 2 a + a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2}),$
$y = - a t_{1} t_{2} (t_{1} + t_{2})$
or $x = 2 a + a [{(t_{1} + t_{2})}_{1}^{2} - t_{1} t_{2}] = 2 a - h + \frac{k^{2}}{a}$
$y = - \frac{h k}{a} .$
Ans: B

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