If the two pairs of the lines $2 x^{2} + 6 x y + y^{2} = 0$ and $4 x^{2} + 18 x y + b y^{2} = 0$ are equally inclined, then b=

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Solution

$2 x^{2} + 6 x y + y^{2} = 0$ & $4 x^{2} + 1 l x y + b y^{2} = 0$
for equakky inclined
$\frac{x^{2} - y^{2}}{2^{2} - 1^{2}} = \frac{x y}{3}$
$x^{2} y^{2} = x y - - - - - (1)$
$\frac{x^{2} - y^{2}}{4^{2} - b^{2}} = \frac{x y}{3}$
$\frac{x y}{16 - b^{2}} = \frac{x y}{3}$
$9 = 16 - b^{2}$
$b^{2} = 7$
$b = \pm \sqrt{7}$

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