If the coordinates of the Parabola of the ends of a focal chord of Parabola $y^{2} = 4 a x$ are $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ , then prove that : $t_{1} . t_{2} = - 1$

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Solution

Let $y^{2} = 4 a x$ be the parabola and $P (a t_{1}^{2}, 2 a t_{1}), Q (a t_{2}^{2}, 2 a t_{2})$ be the endpoints of the focal chord that passes through the focus $S (a, 0)$ .

Then $P S, S Q$ have same slopes.

$\frac{2 a t_{1} - 0}{a t_{1}^{2} - a} = \frac{2 a t_{2} - 0}{a t_{2}^{2} - a}$

$\Rightarrow \frac{2 a t_{1}}{a t_{1}^{2} - a} = \frac{2 a t_{2}}{a t_{2}^{2} - a}$

$\Rightarrow \frac{t_{1}}{t_{1}^{2} - 1} = \frac{t_{2}}{t_{2}^{2} - 1}$

$\Rightarrow t_{1} t_{2}^{2} - t_{1} = t_{2} t_{1}^{2} - t_{2}$

$\Rightarrow t_{2} - t_{1} = t_{2} t_{1}^{2} - t_{1} t_{2}^{2}$

$\Rightarrow t_{2} - t_{1} = t_{1} t_{2} (t_{1} - t_{2})$

$\Rightarrow t_{1} t_{2} = - 1$

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