A chord of the parabola $y^{2} = 4 a x$ subtends a right angle at the vertex. Find the locus of the point of intersection of tangents at its extremities.

x - 4 a = 0.

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x + 2 a = 0.

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x + 4 a = 0.

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x - 2 a = 0.

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Solution

The correct option is B $x + 4 a = 0.$
$A B$ subtends a right angle at the vertex
$∴ t_{1} t_{2} = - 1$ ...(1)
If $(h, k)$ be the point of intersection of tangents at $A$ and $B$ , then
$h = a t_{1} t_{2}, k = a (t_{1} + t_{2})$ or $h = - 4 a$ ................. by (1)
Hence, the locus is $x + 4 a = 0$
Ans: C

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x + 4 a = 0

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t_{1}

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t_{1} t_{2} = - 2

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