For $a, b, h>0$, if the slope of one of the lines represented by $a^2 x^2+2 h x y+b^2 y^2=0$ is twice that…
For $a, b, h>0$, if the slope of one of the lines represented by $a^2 x^2+2 h x y+b^2 y^2=0$ is twice that of the other, then the value of $\frac{h}{a b}$ is
$\frac{3 \sqrt{2}}{4}$
$\frac{2 \sqrt{3}}{4}$
$\frac{-2 \sqrt{3}}{4}$
$\frac{-3 \sqrt{2}}{4}$
Solution
Given, lines are $a^2 x^2+2 h x y+b^2 y^2=0$
Let $m$ and $2 m$ be the slopes of the lines.
Then, $m+2 m=\frac{-2 h}{b^2}$.
$\Rightarrow \quad 3 m=-\frac{2 h}{b^2}$
$\Rightarrow \quad m=-\frac{2 h}{3 b^2}$...(i)
and $m \times 2 m=\frac{a^2}{b^2}$
$\begin{array}{ll}\Rightarrow & 2 m^2=\frac{a^2}{b^2} \\ \Rightarrow & 2\left(-\frac{2 h}{3 b^2}\right)^2=\frac{a^2}{b^2} \quad \text { [from Eq. (i)] }\end{array}$
$\begin{array}{ll}\Rightarrow & \frac{8 h^2}{9 b^4}=\frac{a^2}{b^2} \Rightarrow \frac{h^2}{a^2 b^2}=\frac{9}{8} \\ \Rightarrow & \frac{h}{a b}=\frac{3}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{3 \sqrt{2}}{4}\end{array}$