If the slopes of the lines given by the equat | vedclass

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(D) Let the slopes of the lines be $m_1$ and $m_2$. The equation $ax^2+2hxy+by^2=0$ represents two lines passing through the origin.We have the sum of

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$16:15$

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(D) Let the slopes of the lines be $m_1$ and $m_2$. The equation $ax^2+2hxy+by^2=0$ represents two lines passing through the origin.We have the sum of slopes $m_1+m_2 = -\frac{2h}{b}$ and the product of slopes $m_1m_2 = \frac{a}{b}$.Given the ratio of slopes is $m_1:m_2 = 5:3$,let $m_1 = 5k$ and $m_2 = 3k$.Then $m_1+m_2 = 8k = -\frac{2h}{b} \Rightarrow k = -\frac{h}{4b}$.And $m_1m_2 = 15k^2 = \frac{a}{b}$.Substituting $k$ in the product equation: $15(-\frac{h}{4b})^2 = \frac{a}{b}$.$15 	imes \frac{h^2}{16b^2} = \frac{a}{b}$.$\frac{15h^2}{16b^2} = \frac{a}{b} \Rightarrow \frac{h^2}{ab} = \frac{16}{15}$.Thus,the ratio $h^2:ab$ is $16:15$.

If the slopes of the lines given by the equation $ax^2+2hxy+by^2=0$ are in the ratio $5:3$,then the ratio $h^2:ab$ is:

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