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(A) The given equation of the pair of lines is $ax^2 + (2a + 1)xy + 2y^2 = 0$. Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we have $A = a$,$2H = 2a +

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$\frac{17}{4}$

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(A) The given equation of the pair of lines is $ax^2 + (2a + 1)xy + 2y^2 = 0$. Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we have $A = a$,$2H = 2a + 1$,and $B = 2$. Let the slopes of the lines be $m_1$ and $m_2$. Given that $m_1 = \frac{1}{m_2}$,which implies $m_1 m_2 = 1$. For a pair of lines $ax^2 + 2hxy + by^2 = 0$,the product of slopes is $m_1 m_2 = \frac{A}{B} = \frac{a}{2}$. Equating the two,we get $\frac{a}{2} = 1$,so $a = 2$. The sum of the slopes is $m_1 + m_2 = -\frac{2H}{B} = -\frac{2a + 1}{2}$. Substituting $a = 2$,we get $m_1 + m_2 = -\frac{2(2) + 1}{2} = -\frac{5}{2}$. We need to find $m_1^2 + m_2^2$. Using the identity $m_1^2 + m_2^2 = (m_1 + m_2)^2 - 2m_1 m_2$,we get: $m_1^2 + m_2^2 = (-\frac{5}{2})^2 - 2(1) = \frac{25}{4} - 2 = \frac{25 - 8}{4} = \frac{17}{4}$.

If the slope of one of the lines represented by $ax^2 + (2a + 1)xy + 2y^2 = 0$ is the reciprocal of the slope of the other,then the sum of the squares of the slopes is

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