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Question

If one of the lines given by $6 x^{2} - x y + 4 c y^{2} = 0$ is $3 x + 4 y = 0,$ then $c$ equals

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Solution

It is given that $3 x + 4 y = 0$ i.e. $y = - \frac{3}{4} x$
the lines given by $6 x^{2} - x y + 4 c y^{2} = 0 .$ There, $y = - \frac{3}{4} x$
satisfies it.
$∴ 6 x^{2} + \frac{3}{4} x^{2} + 4 c \times \frac{9}{16} x^{2} = 0 \Rightarrow \frac{27}{4} + \frac{9 c}{4} = 0 \Rightarrow c = - 3$

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