If $\frac{x^2}{\mathrm{a}}+\frac{2 x y}{\mathrm{~h}}+\frac{y^2}{\mathrm{~b}}=0$ represents a pair of straight lines and slope of one of the lines is twice that of the other, then $a b: h^2$ is [Note: The question has been modified to get the correct answer.]

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Aicharya · Accepted Answer

$\mathrm{m}_1: \mathrm{m}_2=1: 2$
If the slopes of the lines given by $\mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2=0$ are in the ratio m:n, then ${ }^{-}$ $(\mathrm{m}+\mathrm{n})^2 \mathrm{ab}=4 \mathrm{mnh}^2$ $\therefore \quad(1+2)^2\left(\frac{1}{a}\right)\left(\frac{1}{b}\right)=4(1)(2)\left(\frac{1}{h}\right)^2 \quad \Rightarrow \frac{a b}{h^2}=\frac{9}{8}$ [Note: In the question, $\frac{x^2}{\mathrm{a}^2}$ is changed to $\frac{x^2}{\mathrm{a}}$ to apply appropriate textual concepts.]

If $\frac{x^2}{\mathrm{a}}+\frac{2 x y}{\mathrm{~h}}+\frac{y^2}{\mathrm{~b}}=0$ represents a pair of…

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