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Secant of a Curve y =f(x)

If P1Q1 P...

Question

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

A

directrix

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B

axis

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C

tangent at the vertex

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D

none of these

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Solution

The correct option is C directrix
Let $P_{1} ({a t_{1}}^{2}, 2 a t_{1}) Q_{1} (\frac{a}{{t_{1}}^{2}}, \frac{- 2 a}{t_{1}})$
$P_{2} ({a t_{2}}^{2}, 2 a t_{2}) Q_{2} (\frac{a}{{t_{2}}^{2}}, \frac{- 2 a}{t_{2}})$
Equation of $P_{1} Q_{2}$
$y - 2 a t_{1} = \frac{\frac{- 2 a}{t_{2}} - 2 a t_{1}}{\frac{a}{t_{2}^{2}} - {a t}_{1}^{2}} (x - a t_{1}^{2}) y - 2 a t_{1} = \frac{\frac{- 2}{t_{2}} - 2 t_{1}}{\frac{1}{t_{2}^{2}} - t_{1}^{2}} (x - a t_{1}^{2}) y - 2 a t_{1} = \frac{- 2 (\frac{1}{t_{2}} + t_{1})}{(\frac{1}{t_{2}} - t_{1}) (\frac{1}{t_{2}} + t_{1})} (x - a t_{1}^{2}) y - 2 a t_{1} = \frac{2}{(\frac{1}{t_{2}} - t_{1})} (x - a t_{1}^{2}) . . . . . . . . . . (1)$
Equation of $P_{2} Q_{1}$
$y - 2 a t_{2} = \frac{\frac{- 2 a}{t_{1}} - 2 a t_{2}}{\frac{a}{t_{1}^{2}} - {a t}_{2}^{2}} (x - a t_{2}^{2}) y - 2 a t_{2} = \frac{2}{(\frac{1}{t_{2}} - t_{2})} (x - a t_{2}^{2}) . . . . . . . . . . (2)$
Solve 1 & 2 [Subtract (2) from (1)]
$2 a (t_{2} - t_{1}) = \frac{2}{t_{2} t_{1} - 1} [t_{2} (x - a t_{1}^{2}) - t_{1} (x - a t_{2}^{2})] a (t_{2} - t_{1}) (t_{2} t_{1} - 1) = (t_{2} - t_{1}) x + a (t_{1} t_{2}^{2} - t_{1}^{2} t_{2}) a (t_{2} - t_{1}) (t_{2} t_{1} - 1) = (t_{2} - t_{1}) [x + a t_{2} t_{1}] a (t_{2} t_{1} - 1) = x + a t_{2} t_{1} x = - a$
Intersection point of $P_{1} Q_{2}$ and $P_{2} Q_{1}$ lie on directrix.

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P_{1} Q_{1}

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P_{1} Q_{1}

P_{2} Q_{2}

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