If the slopes of the lines given by $a x^{2} + 2 h x y + b y^{2} = 0$ are in the ratio $3 : 1$ , then $h^{2}$ is equal to

\frac{a b}{3}

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\frac{4 a b}{3}

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\frac{4 a}{3 b}

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None of these

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Solution

The correct option is B $\frac{4 a b}{3}$
Let $y = m_{1} w$ and $y = m_{2} x$ be the lines represented by the given equation.

Then, $m_{1} + m_{2} = \frac{- 2 h}{b}$ and $m_{1} m_{2} = \frac{a}{b}$

We have, $m_{1} : m_{2} = 3 : 1$

$\Rightarrow m_{1} = 3 m_{2}$

Since $m_{1} + m_{2} = \frac{- 2 h}{b}$ and $m_{1} m_{2} = \frac{a}{b}$

Thus $4 m_{2} = - \frac{2 h}{b}$ and $3 m_{2}^{2} = \frac{a}{b}$

$\Rightarrow m_{2} = \frac{h}{2 b}$ and $m_{2}^{2} = \frac{a}{3 b}$

On solving above equations, we get
$3 {(- \frac{h}{2 b})}^{2} = \frac{a}{b}$
$\Rightarrow h^{2} = \frac{4 a b}{3}$

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