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General Equation of Hyperbola

If P1P2 and...

Question

If $P_{1} P_{2}$ and $P_{3} P_{4}$ are two focal chord of the parabola $y^{2} = 4 a x$ then the chord $P_{1} P_{3}$ and $P_{2} P_{4}$ are intersecting on

A

directrix

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B

vertex

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C

on parabola

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D

not intersect

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Solution

The correct option is B directrix
If $P_{1} P_{2}$ are two focal chord of parabola $y^{2} = 4 a x$ then cord $P_{1} P_{3}$ and $P_{2} P_{4}$ are intersect at :
$P_{1} ({a t}_{1}^{2}, 2 a t_{1}), Q_{2} (\frac{a}{t_{1}^{2}}, \frac{- 2 a}{t_{1}}) P_{1} ({a t}_{2}^{2}, 2 a t_{2}), Q_{2} (\frac{a}{t_{2}^{2}}, \frac{- 2 a}{t_{2}})$
Equation of $P_{1} Q_{2}$
$y - 2 a t_{1} = \frac{\frac{- 2 a}{t_{2}} - 2 a t_{1}}{\frac{a}{t_{2}^{2}} - {a t}_{1}^{2}} (x - {a t}_{1}^{2}) y - 2 a t_{1} = \frac{2}{t_{1} - \frac{1}{t_{2}}} (x - {a t}_{1}^{2}) . . . . . . . . (1)$
Equation of $P_{2} Q_{1}$
$y - 2 a t_{2} = \frac{\frac{- 2 a}{t_{1}} - 2 a t_{2}}{\frac{a}{t_{1}^{2}} - {a t}_{2}^{2}} (x - {a t}_{2}^{2}) y - 2 a t_{2} = \frac{2}{t_{2} - \frac{1}{t_{1}}} (x - {a t}_{2}^{2}) . . . . . . . . (2)$
Solve (1) and (2)
$2 a (t_{2} - t_{1}) = \frac{2}{t_{1} t_{2} - 1} [t_{2} (x - {a t}_{1}^{2}) - t_{1} (x - {a t}_{2}^{2})] a (t_{2} - t_{1}) (t_{1} t_{2} - 1) = (t_{2} - t_{1}) [x + a t_{2} t_{1}] a t_{2} t_{1} - a = x + a t_{2} t_{1} x = - a$
Therefore intersection of point of $P_{1} Q_{2}$ and $P_{2} Q_{1}$ lies on directrix.

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Similar questions

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} P_{2}

Q_{1} Q_{2}

P_{1} Q_{1} & P_{2} Q_{2}

or are two focal chords of

y^{2} = 4 a x,

P_{1} P_{2} & Q_{1} Q_{2}

Q. Assertion :If

P_{1} Q_{1}

P_{2} Q_{2}

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y^{2} = 4 a x

, then the locus of point of intersection of chords

P_{1} P_{2}

Q_{1} Q_{2}

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Q_{1} Q_{2}

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P_{1} Q_{1}

P_{2} Q_{2}

are two focal chords of the parabola

y^{2} = 4 a x,

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are two focal chords of the parabola

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