If $h^2=a b$, then the slopes of lines represented by $a x^2-2 h x y+b y^2=0$ would be in the ratio
If $h^2=a b$, then the slopes of lines represented by $a x^2-2 h x y+b y^2=0$ would be in the ratio
$1: 2$
$2: 1$
$2: 3$
$1: 1$
Solution
Given, $h^2=a b$ and $a x^2+2 h x y+b y^2=0$
Divide by $x^2$, we obtain
$
b m^2+2 h m+a=0
$
$[\because y=m x$ is equation of line through origin, i.e.
$
\left.\frac{y}{x}=m\right]
$
$
\begin{aligned}
\Rightarrow \quad m & =\frac{-2 h \pm \sqrt{4 h^2-4 a b}}{2 b} \\
& =\frac{-2 h \pm \sqrt{4\left(h^2-a b\right)}}{2 b \quad\left(\because h^2=a b \text { and } \therefore h^2-a b=0\right)} \\
& =\frac{-2 h}{2 b} \\
\therefore \quad m_1 & =\frac{h}{b} \text { and } m_2=\frac{h}{b}
\end{aligned}
$
Therefore, slopes of both lines are equal
$
\therefore m_1: m_2=1: 1
$