Two tangents to a parabola intercept on a fixed tangent segments whose product is constant; prove that the locus of their point of intersection is a straight line.

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Solution

Point of intersection of the tangents drawn at the ends of a chord of $y^{2} = 4 a x$ is $R = (a t_{1} t_{2}, a (t_{1} + t_{2}))$
Let the point of contact of variable tangents be $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ and their point of intersection is $(h, k)$

Point of intersection of tangents is $(a t_{1} t_{2}, a (t_{1} + t_{2}))$

$\Rightarrow h = a t_{1} t_{2} . . . . . . . (i) k = a (t_{1} + t_{2}) . . . . . . . (i i)$

Let the point of contact of fixed tangent be $A (a t^{2}, 2 a t)$

Point of intersection of tangent at $P$ and $A$ is $B (a t t_{1}, a (t + t_{1}))$

Point of intersection of tangent at $Q$ and $A$ is $C (a t t_{2}, a (t + t_{2}))$

Given $B A \times A C = c o n s t a n t = c$

$\sqrt{{(a t^{2} - a t t_{1})}^{2} + {(2 a t - a t_{1} - a t)}^{2}} \times \sqrt{{(a t^{2} - a t t_{2})}^{2} + {(2 a t - a t_{2} - a t)}^{2}} = c \sqrt{{a^{2} t^{2} (t - t_{1})}^{2} + {a^{2} (t - t_{1})}^{2}} \times \sqrt{{a^{2} t^{2} (t - t_{2})}^{2} + {a^{2} (t - t_{2})}^{2}} = c a (t - t_{1}) \sqrt{t^{2} + 1} \times a (t - t_{2}) \sqrt{t^{2} + 1} = c (t - t_{1}) (t - t_{2}) = \frac{c}{a^{2} (t^{2} + 1)} (t - t_{1}) (t - t_{2}) = d (s a y)$

$t^{2} - (t_{1} + t_{2}) + t_{1} t_{2} = d$

Substituting $(i)$ and $(i i)$ , we get

$t^{2} - \frac{k}{a} + \frac{h}{a} = d h - k = a (d - t^{2})$

Replacing $h$ by $x$ and $k$ by $y$

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

which represents a straight line.

Hence proved.

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