If the slope of one of the two lines $\frac{x^2}{a}+\frac{2 x y}{h}+\frac{y^2}{b}=0$ is twice that of the other, then $a b: h^2=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \frac{x^2}{a}+\frac{2 x y}{h}+\frac{y^2}{b}=0 \ & \Rightarrow \frac{1}{b}\left(\frac{y}{x}ight)^2+\frac{2}{h}\left(\frac{y}{x}ight)+\frac{1}{a}=0 \ & \Rightarrow m_1+m_2=\frac{\frac{-2}{h}}{\frac{1}{b}}=\frac{-2 b}{h} 	ext { and } m_1 m_2=\frac{\frac{1}{a}}{\frac{1}{b}}=\frac{b}{a}\end{aligned}$ $\begin{aligned} & 	ext { from (1) and } 2\left(\frac{-2 b}{3 h}ight)^2=\frac{b}{a} \ & \Rightarrow 2 	imes \frac{4 b^2}{9 h^2}=\frac{b}{a} \ & \Rightarrow \frac{a b}{h^2}=\frac{9}{8}=9: 8\end{aligned}$

If the slope of one of the two lines $\frac{x^2}{a}+\frac{2 x y}{h}+\frac{y^2}{b}=0$ is twice that of the…

Solution