If the slope of one line of the pair of lines | vedclass

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(C) The equation of the pair of lines is $ax^2 + 4xy + y^2 = 0$.Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we have $A = a$,$2H = 4$ (so $H = 2$),and

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$3$

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(C) The equation of the pair of lines is $ax^2 + 4xy + y^2 = 0$.Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we have $A = a$,$2H = 4$ (so $H = 2$),and $B = 1$.Let the slopes of the two lines be $m_1$ and $m_2$.Then $m_1 + m_2 = -\frac{2H}{B} = -\frac{4}{1} = -4$ $(i)$.And $m_1 m_2 = \frac{A}{B} = \frac{a}{1} = a$ $(ii)$.Given that $m_1 = 3m_2$.Substituting this into $(i)$,we get $3m_2 + m_2 = -4$,which implies $4m_2 = -4$,so $m_2 = -1$.Then $m_1 = 3(-1) = -3$.Substituting these values into $(ii)$,we get $a = m_1 m_2 = (-3)(-1) = 3$.

If the slope of one line of the pair of lines represented by $ax^2 + 4xy + y^2 = 0$ is $3$ times the slope of the other line,then $a$ is

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