A circle drawn on any focal chord the parabola again intersects at $C$ and $D$ . If the parameters of the points $A, B, C, D$ be $t_{1}, t_{2}, t_{3}$ and $t_{4}$ respectively, then the value of $t_{3} t_{4}$ is :

- 1

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2

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3

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none of these.

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Solution

The correct option is C $3$
Since, $A B$ is a focal chord, $t_{1} t_{2} = - 1$ ...(1)
Circle on $A B$ as diameter is
$(x - a t_{1}^{2}) (x - a t_{2}^{2}) + (y - 2 a t_{1}) (y - 2 a t_{2}) = 0$
$x^{2} + y^{2} - a x (t_{1}^{2} + t_{2}^{2}) - 2 a y (t_{1} + t_{2})$
$+ a^{2} t_{1}^{2} t_{2}^{2} + 4 a^{2} t_{1} t_{2} = 0$
It meets the parabola $y^{2} = 4 a x,$ i.e., $x = a T^{2}$
$y = 2 a T$ in four points $A, B, C, D$
$a^{2} T^{4} + 4 a^{2} T^{2} - a^{2} T^{2} (t_{1}^{2} + t_{2}^{2}) - 4 a^{2} T (t_{1} + t_{2}) - 3 a^{2} = 0$
It is a fourth degree equation in $T$ whose roots are
$t_{1}, t_{2}, t_{3}, t_{4}$
$∴ t_{1} . t_{2} . t_{3} . t_{4} = - \frac{3 a^{2}}{a^{2}} = - 3.$ Put $t_{1} . t_{2} = - 1$
$∴ t_{3} . t_{4} = 3$

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