Find $k$ , if the sum of slopes of lines represented by equation $x^{2} + k x y - 3 y^{2} = 0$ is twice their product.

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Solution

Let the lines represented by the equation.
$x^{2} + k x y - 3 y^{2} = 0$

$\Rightarrow$ $- \frac{x^{2}}{3} - \frac{k}{3} x y + y^{2} = 0$ be $y = m_{1} x$ and $y = m_{2} x$ , where $m_{1}$ and $m_{2}$ are the slopes of the lines.

$\Rightarrow$ $- \frac{x^{2}}{3} - \frac{k}{3} x y + y^{2} = (y - m_{1} x) (y - m_{2} x)$

$\Rightarrow$ $- \frac{x^{2}}{3} - \frac{k}{3} x y + y^{2} = y^{2} - (m_{1} + m_{2}) x y + (m_{1} m_{2}) x^{2}$

Comparing both sides, we get
$C (m_{1} + m_{2}) = \frac{k}{3}$ and $(m_{1} m_{2}) = - \frac{1}{3}$

Given, $m_{1} + m_{2} = 2 m_{1} m_{2}$

$∴ \frac{k}{3} = 2 (- \frac{1}{3})$

$∴ k = - 2$

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