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Double Ordinate

If A1 B1 an...

Question

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

A

directrix

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B

axis

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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
$A_{1} B_{1}$ is a focal chord, then $A_{1} (a t_{1}^{2}, 2 a t_{1})$ and $B_{1} (\frac{a}{t_{1}^{2}}, \frac{- 2 a}{t_{1}})$ .
$A_{2} B_{2}$ is a focal chord, then $A_{2} (a t_{2}^{2}, 2 a t_{2})$ and $B_{2} (\frac{a}{t_{2}^{2}}, \frac{- 2 a}{t_{2}})$
Now equation of chord $A_{1} A_{2}$ is
$y - 2 a t_{1} = \frac{2}{t_{1} + t_{2}} (x - a {t_{1}}^{2})$
$y (t_{1} + t_{2}) - 2 x - 2 a t_{1} t_{2} = 0$ ..... (i)
Chord $B_{1} B_{2}$ is
Chord $B_{1} B_{2}$ is
$y (- \frac{1}{t_{1}} - \frac{1}{t_{2}}) - 2 x - 2 a (- \frac{1}{t_{1}}) (- \frac{1}{t_{2}}) = 0$
or $y (t_{1} + t_{2}) + 2 x t_{1} t_{2} + 2 a = 0$ .....(ii)
Solving (i) and (ii), we get
$2 x (t_{1} t_{2} + 1) + 2 a (t_{1} t_{2} + 1) = 0$
or $(x + a) (1 + t_{1} t_{2}) = 0$
$\Rightarrow x + a = 0$
Hence, they intersect on directrix.

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