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Two Point Form of a Line

If t1 and ...

Question

If $t_{1}$ and $t_{2}$ are the roots ofthe equation $t^{2} + λ t + 1 = 0$ . Where, $λ$ is an aribitary constant. Then the line joining the points $(a {t_{1}}^{2}, 2 a t_{1})$ and $(a {t_{2}}^{2}, 2 a t_{2})$ passes through a fixed point, which is

A

(a, 0)

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B

(- a, 0)

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C

(0, a)

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D

(0, - a)

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Solution

The correct option is A $(- a, 0)$
Then equation of line joining the points $(a {t_{1}}^{2}, 2 a t_{1})$ and $(a {t_{2}}^{2}, 2 a t_{2})$ is given by,
$(y - 2 a t_{1}) = \frac{2 a t_{1} - 2 a t_{2}}{a t_{1}^{2} - a t_{2}^{2}} (x - a t_{1}^{2})$

$\Rightarrow y - 2 a t_{1} = \frac{2}{t_{1} + t_{2}} (x - a t_{1}^{2}) \Rightarrow (1)$
Now since $t_{1}, t_{2}$ are the roots of quadratic $t^{2} + λ t + 1 = 0$
$∴ t_{1} t_{2} = 1 \Rightarrow (2)$
Thus using (2) in (1) we get,
$\Rightarrow y - 2 a t_{1} = \frac{2}{t_{1} + 1 / t_{1}} (x - a t_{1}^{2})$
$\Rightarrow y = \frac{2 t_{1}}{1 + t_{1}^{2}} (x - a t_{1}^{2}) + 2 a t_{1} = \frac{2 t_{1}}{1 + t_{1}^{2}} (x - a t_{1}^{2}) + \frac{2 a t_{1} (1 + t_{1}^{2})}{1 + t_{1}^{2}}$

$\Rightarrow y = \frac{2 t_{1}}{1 + t_{1}^{2}} (x - a t_{1}^{2} + a + a t_{1}^{2}) = \frac{2 t_{1}}{1 + t_{1}^{2}} (x + a)$
Put $\frac{2 t_{1}}{1 + t_{1}^{2}} = k$
Thus the line becomes, $y = k (x + a)$
Clearly this line is passing through a fixed point $(- a, 0)$

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