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(A) The given equation is $ax^2+8xy+5y^2=0$. Comparing this with the general equation $Ax^2+2Hxy+By^2=0$,we have $A=a$,$2H=8$ (so $H=4$),and $B=5$. Le

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-$4$

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(A) The given equation is $ax^2+8xy+5y^2=0$. Comparing this with the general equation $Ax^2+2Hxy+By^2=0$,we have $A=a$,$2H=8$ (so $H=4$),and $B=5$. Let the slopes of the lines be $m_1$ and $m_2$. The sum of slopes is $m_1+m_2 = -\frac{2H}{B} = -\frac{8}{5}$. The product of slopes is $m_1m_2 = \frac{A}{B} = \frac{a}{5}$. According to the given condition,the sum of slopes is twice their product: $m_1+m_2 = 2(m_1m_2)$ $-\frac{8}{5} = 2\left(\frac{a}{5}\right)$ $-\frac{8}{5} = \frac{2a}{5}$ $2a = -8$ $a = -4$.

If the sum of slopes of lines represented by $ax^2+8xy+5y^2=0$ is twice their product,then $a=$

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