2015_22 KVPY 2015 Preparations-2

Geometry Level 4

If the slope of one of the lines represented by 4 a x 2 + x y + 4 y 2 = 0 4a{ x }^{ 2 }+xy+4{ y }^{ 2 }=0 is the square of the other, then what is the value of a a ?

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The answer is -0.125.

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1 solution

Tapas Mazumdar
Apr 5, 2017

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 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 m_1 = {m_2}^2 .

Now, we have

4 a x 2 + x y + 4 y 2 = ( y m 1 x ) ( y m 2 x ) = y 2 + x y ( m 1 m 2 ) + m 1 m 2 x 2 = y 2 + x y ( m 2 2 m 2 ) + m 2 3 x 2 \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.,

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

Now

1 ( m 2 2 + m 2 ) = 4 1 = 4 m 2 2 4 m 2 4 m 2 2 + 4 m 2 + 1 = 0 ( 2 m 2 + 1 ) 2 = 0 m 2 = 1 2 \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

4 a m 2 3 = 4 a = m 2 3 a = ( 1 2 ) 3 = 1 8 = 0.125 \dfrac{4a}{{m_2}^3} = 4 \implies a = {m_2}^3 \implies a = {\left( - \dfrac 12 \right)}^3 = - \dfrac 18 = \boxed{-0.125}

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