If the slopes of the lines given by the equation $a x^2+2 h x y+b y^2=0$ are in the ratio $5: 3$, then ration $h^2: a b=$

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Aicharya · Accepted Answer

Let $y=m_1 x$ and $y=m_2 x$ be the lines represented by the equation. $a x^2+2 h x y+b y^2=0$ Then, $\mathrm{m}_1+\mathrm{m}_2=\frac{-2 \mathrm{~h}}{\mathrm{~b}}$ and $\mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{a}}{\mathrm{b}}$ We have, $\frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{5}{3} \Rightarrow \mathrm{m}_1=\frac{5 \mathrm{~m}_2}{3}$ $\begin{aligned} & 	herefore \frac{5 \mathrm{~m}_2}{3}+\mathrm{m}_2=\frac{-2 \mathrm{~h}}{\mathrm{~b}} 	ext { and }\left(\frac{5 \mathrm{~m}_2}{3}ight) \mathrm{m}_2=\frac{\mathrm{a}}{\mathrm{b}} \ & 	herefore \frac{8 \mathrm{~m}_2}{3}=\frac{-2 \mathrm{~h}}{\mathrm{~b}} \Rightarrow \mathrm{m}_2=\frac{-3 \mathrm{~h}}{4 \mathrm{~b}} 	ext { and } \mathrm{m}_2^2=\frac{3 \mathrm{a}}{5 \mathrm{~b}} \ & \left(\frac{-3 \mathrm{~h}}{4 \mathrm{~b}}ight)^2=\frac{3 \mathrm{a}}{5 \mathrm{~b}} \Rightarrow \frac{9 \mathrm{~h}^2}{16 \mathrm{~b}^2}=\frac{3 \mathrm{a}}{5 \mathrm{~b}} \ & 	herefore \frac{\mathrm{h}^2}{\mathrm{ab}}=\frac{16}{15} \end{aligned}$

If the slopes of the lines given by the equation $a x^2+2 h x y+b y^2=0$ are in the ratio $5: 3$, then…

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