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Equation of Parabola When Its Axis Is Parallel to X or Y Axis

Circles are d...

Question

Circles are drawn taking any two focal chords of the parabola $y^{2} = 4 a x$ as diameters. If $t_{1}, t_{3}$ represent any one end of the different focal chords, then, the equation of the common chord of the circles is

x (t_{3} + t_{1}) (t_{1}^{2} t_{3}^{2} - 1) + 2 y t_{1} t_{3} (t_{1} t_{3} + 1) = 0

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x (t_{3} + t_{1}) (t_{1}^{2} t_{3}^{2} + 1) = 2 y t_{1} t_{3} (t_{1} t_{3} + 1)

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x (t_{3} - t_{1}) (t_{1}^{2} t_{3}^{2} + 1) = 2 y t_{1} t_{3} (t_{1} t_{3} + 1)

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x (t_{3} - t_{1}) (t_{1}^{2} t_{3}^{2} + 1) + 2 y t_{1} t_{3} (t_{1} t_{3} + 1) = 0

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Solution

The correct option is A $x (t_{3} + t_{1}) (t_{1}^{2} t_{3}^{2} - 1) + 2 y t_{1} t_{3} (t_{1} t_{3} + 1) = 0$
Let $P Q$ and $R S$ be two focal chord of the given parabola and let $P, Q, R, S$ be the points $t_{1}, t_{2}, t_{3}, t_{4}$ respectively, then
$t_{1} t_{2} = - 1$ and $t_{3} t_{4} = - 1$
As per the question $P Q$ is a diameter for a perticular circle, then the equation of the circle is
$(x - a t_{1}^{2}) (x - a t_{2}^{2}) + (y - 2 a t_{1}) (y - 2 a t_{2}) = 0 \Rightarrow x^{2} + y^{2} - a (t_{1}^{2} + t_{2}^{2}) x - 2 a (t_{1} + t_{2}) y - 3 a^{2} = 0 \dots (1)$
and circle with the diameter as $R S$ will be
$x^{2} + y^{2} - a (t_{3}^{2} + t_{4}^{2}) x - 2 a (t_{3} + t_{4}) y - 3 a^{2} = 0 \dots (2)$
Hence the equation of common chord will be
$a x (t_{3}^{2} + t_{4}^{2} - t_{1}^{2} - t_{2}^{2}) + 2 a y (t_{3} + t_{4} - t_{1} - t_{2}) = 0 \Rightarrow a x ((t_{3}^{2} - t_{1}^{2}) + \frac{1}{t_{3}^{2}} - \frac{1}{t_{1}^{2}}) + 2 a y ((t_{3} - t_{1}) + \frac{1}{t_{1}} - \frac{1}{t_{3}}) = 0 \Rightarrow a x (t_{3}^{2} - t_{1}^{2}) (t_{1}^{2} t_{3}^{2} - 1) + 2 a y (t_{3} - t_{1}) t_{1} t_{3} (t_{1} t_{3} + 1) = 0$
as $t_{1} \neq t_{3}$
$∴ x (t_{3} + t_{1}) (t_{1}^{2} t_{3}^{2} - 1) + 2 y t_{1} t_{3} (t_{1} t_{3} + 1) = 0$

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