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(B) The given equation of the pair of lines is $3x^2 - 2xy - 15y^2 = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we get $a =

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$2:3$

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(B) The given equation of the pair of lines is $3x^2 - 2xy - 15y^2 = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we get $a = 3$,$2h = -2$ (so $h = -1$),and $b = -15$. Let the slopes of the lines be $m_1$ and $m_2$. The sum of the slopes is $s = m_1 + m_2 = \frac{-2h}{b} = \frac{-2(-1)}{-15} = \frac{2}{-15} = -\frac{2}{15}$. The product of the slopes is $p = m_1 m_2 = \frac{a}{b} = \frac{3}{-15} = -\frac{3}{15}$. Therefore,the ratio $s:p = \left(-\frac{2}{15}\right) : \left(-\frac{3}{15}\right) = 2:3$.

If $s$ and $p$ are respectively the sum and the product of the slopes of the lines $3x^2 - 2xy - 15y^2 = 0$,then $s:p$ is equal to

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