If 4ab = 3h^2, then the ratio of the slopes o | vedclass

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(D) Let the slopes of the lines be $m_1$ and $m_2$. For the equation $ax^2 + 2hxy + by^2 = 0$, we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \fra

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$1 : 3$

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(D) Let the slopes of the lines be $m_1$ and $m_2$. For the equation $ax^2 + 2hxy + by^2 = 0$, we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \frac{a}{b}$.The difference of the slopes is $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1 m_2} = \sqrt{\frac{4h^2}{b^2} - \frac{4a}{b}} = \sqrt{\frac{4h^2 - 4ab}{b^2}}$.Given $4ab = 3h^2$, substitute this into the expression: $|m_1 - m_2| = \sqrt{\frac{4h^2 - 3h^2}{b^2}} = \sqrt{\frac{h^2}{b^2}} = \left| \frac{h}{b} \right|$.Now, consider the ratio $m_1 : m_2$. From $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 - m_2 = \frac{h}{b}$, adding gives $2m_1 = -\frac{h}{b} \implies m_1 = -\frac{h}{2b}$.Subtracting gives $2m_2 = -\frac{3h}{b} \implies m_2 = -\frac{3h}{2b}$.The ratio $m_1 : m_2 = \left( -\frac{h}{2b} \right) : \left( -\frac{3h}{2b} \right) = 1 : 3$ or $3 : 1$.

List-$I$	List-$II$
$(A)$ $5x^2 + 2\sqrt{7}xy - y^2 = 0$	$(I)$ $\frac{\sqrt{3}}{2}$
$(B)$ $x^2 + \sqrt{11}xy + 2y^2 = 0$	$(II)$ $\frac{1}{2\sqrt{3}}$
$(C)$ $x^2 + 2\sqrt{2}xy + y^2 = 0$	$(III)$ $\frac{1}{2}$
$(D)$ $3x^2 + 4\sqrt{2}xy + y^2 = 0$	$(IV)$ $\frac{2}{3}$
	$(V)$ $\frac{1}{\sqrt{2}}$

If $4ab = 3h^2$, then the ratio of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is...

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