2015_22 KVPY 2015 Preparations-2

Since we have a homogenous second degree equation representing a pair of straight lines, therefore the constituent lines must both pass through origin. Let the equations of these lines be

$y = m_1 x \\ y = m_2 x$

From the constraint given in the problem, WLOG we can assume that $m_1 = {m_2}^2$ .

Now, we have

$\begin{aligned} 4ax^2 + xy + 4y^2 &= (y-m_1 x)(y-m_2 x) \\ &= y^2 + xy(-m_1-m_2) + m_1 m_2 x^2 \\ &= y^2 + xy(-{m_2}^2-m_2) + {m_2}^3 x^2 \end{aligned}$

Since both these equations represent the same equation of a pair of straight lines, therefore, their coefficients must be in equal ratio, i.e.,

$\dfrac{4a}{{m_2}^3} = \dfrac{1}{-({m_2}^2 + m_2)} = \dfrac{4}{1}$

Now

$\begin{aligned} \dfrac{1}{-({m_2}^2 + m_2)} = 4 & \implies 1 = -4 {m_2}^2 - 4 m_2 \\ & \implies 4{m_2}^2 + 4m_2 +1 = 0 \\ & \implies {(2m_2 + 1)}^2 = 0 \\ & \implies m_2 = - \dfrac 12 \end{aligned}$

And

$\dfrac{4a}{{m_2}^3} = 4 \implies a = {m_2}^3 \implies a = {\left( - \dfrac 12 \right)}^3 = - \dfrac 18 = \boxed{-0.125}$