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

If P 1 Q 1 ...

Question

If $P_{1} Q_{1} & P_{2} Q_{2}$ or are two focal chords of $y^{2} = 4 a x,$ then the chord $P_{1} P_{2} & Q_{1} Q_{2}$ intersect on

A

axis

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B

directrix

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C

tangent at vertex

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D

None of these

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Solution

The correct option is A directrix
Let $P_{1} = (a t_{1}^{2}, 2 a t_{1})$ , $Q_{1} = (\frac{a}{t_{1}^{2}}, - 2 a t_{1})$ and $P_{2} = (a t_{2}^{2}, 2 a t_{2})$ ,
$Q_{2} = (\frac{a}{t_{2}^{2}}, \frac{- 2 a}{t_{2}})$ be the coordinates of the focal chord of the parabola $y^{2} = 4 a x$
Equation of the chord $P_{1} P_{2}$ is
$y (t_{1} + t_{2}) = 2 x + 2 a t_{1} t_{2}$
$∴ y (t_{1} + t_{2}) - 2 x - 2 a t_{1} t_{2} = 0$ ....... $(1)$
Equation of the chord $Q_{1} Q_{2}$ is
$y (\frac{- 1}{t_{1}} + ({\frac{- 1}{t}}_{2})) = 2 x + 2 a \frac{- 1}{t_{1}} \frac{- 1}{t_{2}}$
$- y (\frac{t_{1} + t_{2}}{t_{1} t_{2}}) - 2 x - \frac{2 a}{t_{1} t_{2}}$
$∴ y (t_{1} + t_{2}) + 2 x t_{1} t_{2} + 2 a = 0$ ......... $(2)$
Subtracting eqn $(2)$ from eqn $(1)$ we get
$2 x (t_{1} t_{2} + 1) + 2 a (t_{1} t_{2} + 1) = 0$
$\Rightarrow 2 (x + a) (t_{1} t_{2} + 1) = 0$
$∴ x + a = 0$ is the directrix of the parabola where the line $P_{1} P_{2}$ and $Q_{1} Q_{2}$ intersect.

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