Question

Equation $\lambda x^3 -10x^2y -xy^2 +4y^3 =0$ represented three straight lines, out of these three, two lines makes equal angle with y = x and $\lambda > 0$, then the value of $\lambda$ is

• A. 7
• B. -8
• C. 8
• D. 9

Answer 1

Answer: (A)

7

Answer 2

The given equation is $\lambda x^3 -10x^2y -xy^2 +4y^3 =0$. This is a homogeneous equation of degree 3, representing three straight lines passing through the origin.

To find the slopes of these lines, we can divide the equation by $x^3$ and let $m = y/x$: $4m^3 - m^2 - 10m + \lambda = 0$

Let the roots of this cubic equation be $m_1, m_2, m_3$. These are the slopes of the three lines. From Vieta's formulas:

1. $m_1 + m_2 + m_3 = 1/4$
2. $m_1 m_2 + m_2 m_3 + m_3 m_1 = -10/4 = -5/2$
3. $m_1 m_2 m_3 = -\lambda/4$

The line $y=x$ has a slope $m_L = 1$. The condition that two lines make equal angles with $y=x$ implies that for at least one pair of slopes $(m_i, m_j)$, we must have either $m_i = m_j$ or $m_i m_j = 1$.

Case 1: $m_1 = m_2 = m$. From Vieta's formulas: $2m + m_3 = 1/4$ and $m^2 + 2m m_3 = -5/2$. Substituting $m_3 = 1/4 - 2m$ into the second equation gives $6m^2 - m - 5 = 0$, which factors as $(6m+5)(m-1) = 0$.

• If $m=1$, then $m_1 = m_2 = 1$, and $m_3 = 1/4 - 2(1) = -7/4$. The slopes are $\{1, 1, -7/4\}$. Then $m_1 m_2 m_3 = (1)(1)(-7/4) = -7/4$. Since $m_1 m_2 m_3 = -\lambda/4$, we have $-7/4 = -\lambda/4 \implies \lambda = 7$. This satisfies $\lambda > 0$.
• If $m=-5/6$, then $m_1 = m_2 = -5/6$, and $m_3 = 1/4 - 2(-5/6) = 23/12$. The slopes are $\{-5/6, -5/6, 23/12\}$. Then $m_1 m_2 m_3 = (-5/6)(-5/6)(23/12) = 575/432$. Since $m_1 m_2 m_3 = -\lambda/4$, we have $575/432 = -\lambda/4 \implies \lambda = -2300/432$, which is negative and rejected.

Case 2: $m_1 m_2 = 1$. From Vieta's formulas: $m_1 + m_2 + m_3 = 1/4$ and $m_1 m_2 + m_3(m_1 + m_2) = -5/2$. Substituting $m_1 m_2 = 1$ into the second equation gives $1 + m_3(m_1 + m_2) = -5/2$, so $m_3(m_1 + m_2) = -7/2$. Using $m_1 + m_2 = 1/4 - m_3$, we get $m_3(1/4 - m_3) = -7/2$, which leads to $4m_3^2 - m_3 - 14 = 0$. The solutions for $m_3$ are $2$ and $-7/4$.

• If $m_3 = 2$, then $m_1 + m_2 = 1/4 - 2 = -7/4$. With $m_1 m_2 = 1$, the quadratic $t^2 + (7/4)t + 1 = 0$ has complex roots, which are not valid slopes.
• If $m_3 = -7/4$, then $m_1 + m_2 = 1/4 - (-7/4) = 2$. With $m_1 m_2 = 1$, the quadratic $t^2 - 2t + 1 = 0$ gives $m_1 = m_2 = 1$. The slopes are $\{1, 1, -7/4\}$, which is the same as in Case 1, leading to $\lambda = 7$.

The only valid value for $\lambda$ satisfying $\lambda > 0$ is 7.