<s>[INST] If one of the lines given by $6x^2 – xy + 4cy^2 = 0$ is $3x+4y = 0$ then c is equal to? If one of the lines given by $6x^2 – xy + 4cy^2 = 0$ is $3x+4y = 0,$ then $c$ is equal to?
I have tried solving it by putting $x = \frac{-4y}{3}$ inside the given equation, as $3x + 4y$ is a solution of $6x^2 – xy + 4cy^2 = 0$, but I am getting two variables, $c$ and $y$. Please tell me how to solve it.
 [/INST] Your idea is good; plugging in $x=-\tfrac43y$ yields
$$0=\frac{96}{9}y^2+\frac43y^2+4cy^2=\frac49\left(36+12c\right)y^2,$$
which must hold for all $y$. Then $36+12c=0$ so $c=-3$.
Alternatively, you could note that the point $(x,y)=(4,-3)$ is on the line, and hence on the curve, so
$$6\cdot4^2-4\cdot(-3)+4\cdot c\cdot(-3)^2=0.$$
This shows that $24+3+9c=0$ so $c=-3$.
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