The slopes of the lines given by $x^2+2 h x y+2 y^2=0$ are in the ratio $1: 2$, then $h$ is

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Aicharya · Accepted Answer

$\begin{array}{ll} & x^2+2 h x y+2 y^2=0 \\ \therefore \quad & A=1, B=2, H=h \end{array}$
Given equation of pair of lines is
Let the slopes of the lines be $\mathrm{m}_1, \mathrm{~m}_2$. $\therefore \quad m_1+m_2=\frac{-2 h}{2}, m_1 m_2=\frac{1}{2}$
Given $\frac{m_1}{m_2}=\frac{1}{2}$ $\begin{aligned} & \Rightarrow \mathrm{m}_1=\frac{1}{2} \mathrm{~m}_2 \\ & \Rightarrow \frac{1}{2} \mathrm{~m}_2 \times \mathrm{m}_2=\frac{1}{2} \\ & \Rightarrow \mathrm{~m}_2^2=1 \\ & \Rightarrow \mathrm{~m}_2= \pm 1 \end{aligned}$ $\therefore \quad \mathrm{m}_1= \pm \frac{1}{2} .$
Now, $\mathrm{m}_1+\mathrm{m}_2=\frac{-2 \mathrm{~h}}{2}$ $\begin{array}{ll}\therefore \quad & \text { when } m_1=\frac{1}{2}, m_2=1 \\ & m_1+m_2=\frac{-2 h}{2} \\ \therefore \quad & h=\frac{-3}{2} \\ & \text { When } m_1=\frac{-1}{2}, m_2=-1 \\ \therefore \quad & m_1+m_2=\frac{-2 h}{2} \\ & \frac{-1}{2}-1=-\frac{2 h}{2} \\ \therefore \quad & h=\frac{3}{2}\end{array}$

The slopes of the lines given by $x^2+2 h x y+2 y^2=0$ are in the ratio $1: 2$, then $h$ is

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