If $4 a b=3 h^2$, then the ratio of the slope of lines represented by $a x^2+2 h x y+b y^2=0$ is
If $4 a b=3 h^2$, then the ratio of the slope of lines represented by $a x^2+2 h x y+b y^2=0$ is
- $\sqrt{3}: 1$
- $1: \sqrt{3}$
- $1: 3$
- $3: 1$
Solution
$\begin{aligned}
& \text {Here, } m_1+m_2=\frac{-2 h}{b} ...(i) \\
& \text { and } \mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{a}}{\mathrm{~b}}. \\
& \therefore \quad\left(m_1-m_2\right)^2=\left(m_1+m_2\right)^2-4 m_1 m_2 \\
& =\frac{4 h^2-4 a b}{b^2} \\
& =\frac{4 h^2-3 h^2}{b^2} \ldots\left[\because 4 a b=3 h^2 \text { (given) }\right] \\
& =\frac{h^2}{b^2}
\end{aligned}$
$\therefore \quad \mathrm{m}_1-\mathrm{m}_2=\frac{\mathrm{h}}{\mathrm{~b}}...(ii)$
On solving (i) and (ii), we get
$\begin{aligned}
& \quad m_1=\frac{-h}{2 b} \text { and } m_2=\frac{-3 h}{2 b} \\
& \therefore \quad m_1: m_2=1: 3
\end{aligned}$
Asked in: MHT CET 2024 (04 May Shift 1)
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