lf the pairs of lines $x^{2} + 2 x y + a y^{2} = 0$ and $a x^{2} + 2 x y + y^{2} = 0$ have exactly one line in common then joint equation of the other two lines are given by

3 x^{2} + 8 x y - 3 y^{2} = 0

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3 x^{2} + 10 x y + 3 y^{2} = 0

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y^{2} + 2 x y - 3 x^{2} = 0

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x^{2} + 2 x y - 3 y^{2} = 0

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Solution

The correct option is B $3 x^{2} + 10 x y + 3 y^{2} = 0$
$y = m x$ and $y = m_{1} n$ this two lines give
$x^{2} + 2 x y + a y^{2} = 0$
$m + m_{1} = \frac{- 2}{a}$ ---(1)
$m m_{1} = \frac{1}{a}$ ----(2)
and $y = m x$ and $y = \frac{- 1}{m_{1}} x$ this two lines give
$a x^{2} + 2 x y + y^{2} = 0$
$m - \frac{1}{m_{1}} = - 2$ ---(3)
$\frac{+ m}{m_{1}} = a$ ----(4)
from 2 & 4,
$m_{1}^{2} = \frac{- 1}{a^{2}}$
$m_{1} = \frac{- 2}{a (a - 1)}$ and $m = \frac{- 2}{(a - 1)}$
$(y - m_{1} x) (y + \frac{x}{m_{1}}) = 0$ $\frac{+ 4}{(a - 1)^{2}} = 1$
$y^{2} + (\frac{1}{m} - m_{1}) x y - x^{2} = 0$ $(a - 1)^{2} = 4$
$y^{2} + \frac{x y}{3} - 3 x y - x^{2} = 0$ $a - 1 = 2, - 2$
$3 y^{2} - 8 x y - 3 x^{2} = 0$ $a = 3, - 1$
$3 x^{2} + 8 x y - 3 y^{2} = 0$

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