The slopes of the lines represented by $x^2+2 h x y+2 y^2=0$ are in the ratio $(1: 2)$, then $h$ is equal
The slopes of the lines represented by $x^2+2 h x y+2 y^2=0$ are in the ratio $(1: 2)$, then $h$ is equal
$\pm \frac{1}{2}$
$\pm \frac{3}{2}$
$\pm 1$
$\pm 3$
Solution
Given, equation of lines
$x^2+2 h x y+2 y^2=0...(i)$
On comparing with $a x^2+2 h x y+b y^2=0$
$\Rightarrow \quad a=1$ and $b=2$
Let $y=m_1 x$ and $y=m_2 x$ be the lines, whose combined form is Eq. (i)
$\therefore \quad m_1+m_2=\frac{-2 h}{b}=\frac{-2 h}{2}=-h...(ii)$
and $\quad m_1 m_2=a / b=1 / 2...(iii)$
$\begin{aligned} & \because \quad m_1: m_2=1: 2 \\ & \Rightarrow \quad \frac{m_1}{m_2}=\frac{1}{2} \Rightarrow m_2=2 m_1\end{aligned}$
From Eq. (ii), we get $m_1+m_2=-h$ and from Eq. (iii) we get
$\begin{array}{llrl}m_1+2 m_1=-h & \text { and } & m_1 m_2 & =\frac{1}{2} \\ m_1=-h / 3 & \text { and } & 2 m_1^2 & =\frac{1}{2} \\ \Rightarrow-\frac{h}{3}= \pm \frac{1}{2} \Rightarrow h= \pm \frac{3}{2} & \text { and } & m_1 & =\frac{ \pm 1}{2}\end{array}$