If $t_{1}$ and $t_{2}$ are the extremities of any focal chord of the parabola $y^{2} = 4 a x$ then $t_{1} t_{2} =$

- 1

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0

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\pm 1

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\frac{1}{2}

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Solution

The correct option is D $- 1$
Coordinates of end point of focal chord are $(a t_{1}^{2}, 2 a t_{1}), (a t_{2}^{2}, 2 a t_{2})$ and the focus is $(a, 0)$
Three point are collinear , so
Slopes will be same,
$\frac{2 a t_{2} - 2 a t_{1}}{a t_{2}^{2} - a t_{1}^{2}} = \frac{2 a t_{2} - 0}{a t_{2}^{2} - a}$
$\Rightarrow \frac{t_{2} - t_{1}}{t_{2}^{2} - t_{1}^{2}} = \frac{t_{2}}{t_{2}^{2} - 1}$
$\Rightarrow \frac{t_{2} - t_{1}}{(t_{2} - t_{1}) (t_{2} + t_{1})} = \frac{t_{2}}{t_{2}^{2} - 1}$
$\Rightarrow \frac{1}{t_{2} + t_{1}} = \frac{t_{2}}{t_{2}^{2} - 1}$
$\Rightarrow t_{2}^{2} - 1 = t_{2} (t_{1} + t_{2}) = t_{1} t_{2} + t_{2}^{2}$
$\Rightarrow t_{1} t_{2} = - 1$

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Similar questions

If $t_{1} a n d t_{2}$ are end of a focal chord of the parabola $(y^{2} = 4 a x)$ then $t_{1} . t_{2} = - 1$

y^{2} = 4 a x

(a t_{1}^{2}, 2 a t_{1})

(a t_{2}^{2}, 2 a t_{2})

t_{1} . t_{2} = - 1

y^{2} = 4 a x

A B

A (a t_{1}^{2}, 2 a t_{1})

B (a t_{2}^{2}, 2 a t_{2})