If $t_{1}$ and $t_{2}$ are roots of the equation $t^{2} + \lambda t + 1 = 0,\$where λ is an arbitrary constant. Then, the line joining the points $({at}_{1}^{2},2at_{1})$ and $({at}_{2}^{2},\ 2at_{2})$ always passes through a fixed point whose coordinates are

Question

Dr. Abubacker · Accepted Answer

(b)
Since, $t_{1},\ t_{2}$ are the roots of the equation $t^{2} + \lambda t + 1 = 0$
$	herefore t_{1} + t_{2} = - \lambda,\ \ t_{1}t_{2} = 1$
The equation of a line passing through $({at}_{1}^{2},\ 2at_{1})$ and $({at}_{2}^{2},\ 2at_{2})$ is
$y - 2at_{2} = \frac{2}{t_{1} + t_{2}}\left( x - {at}_{2}^{2} ight)$
$\Rightarrow \ y - 2at_{2} = - \frac{2}{\lambda}\left( x - {at}_{2}^{2} ight)$
$\Rightarrow \ \lambda y - 2a\lambda t_{2} = - 2x + 2{at}_{2}^{2}$
$\Rightarrow \ \lambda y + 2x = 2a\left( \lambda t_{2} + t_{2}^{2} ight)$
$\Rightarrow \ \lambda y + 2x = 2a( - 1)$
$\Rightarrow 2(x + a) + \lambda y = 0$
$	herefore\$Fixed point is $( - a,\ 0)$

Q. If $t_{1}$ and $t_{2}$ are roots of the equation $t^{2} + \lambda t + 1 = 0,\$where λ is an arbitrary constant. Then, the line joining the points $({at}_{1}^{2},2at_{1})$ and $({at}_{2}^{2},\ 2at_{2})$ always passes through a fixed point whose coordinates are

Explanation

Reviewed by Dr. Abubacker

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