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(B) Let the slopes of the lines be $m_{1}$ and $m_{2}$.Given the equation $ax^{2} + 2hxy + by^{2} = 0$,we have $m_{1} + m_{2} = \frac{-2h}{b}$ and $m_

Vedclass · Accepted Answer

$16:15$

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(B) Let the slopes of the lines be $m_{1}$ and $m_{2}$.Given the equation $ax^{2} + 2hxy + by^{2} = 0$,we have $m_{1} + m_{2} = \frac{-2h}{b}$ and $m_{1}m_{2} = \frac{a}{b}$.Given the ratio of slopes is $m_{1}:m_{2} = 5:3$,let $m_{1} = 5k$ and $m_{2} = 3k$.Then $m_{1} + m_{2} = 8k = \frac{-2h}{b} \Rightarrow k = \frac{-h}{4b}$.Also $m_{1}m_{2} = 15k^{2} = \frac{a}{b}$.Substituting $k$ in the second equation: $15 \left( \frac{-h}{4b} \right)^{2} = \frac{a}{b}$.$15 \left( \frac{h^{2}}{16b^{2}} \right) = \frac{a}{b}$.$\frac{15h^{2}}{16b} = a$.Therefore,$\frac{h^{2}}{ab} = \frac{16}{15}$.

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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