Question

For what values of \( \lambda \) does the equation \( 6x^2 - xy + \lambda y^2 = 0 \) represent two perpendicular lines and two lines inclined at an angle of 45°?

• A. \( -6 \) and \( -2 \)
• B. \( 6 \) and \( 1 \)
• C. \( -6 \) and \( -35 \)
• D. \( -6 \) and \( 1 \)

Hint:

For perpendicular lines, use the condition \( B^2 - AC = 0 \), and for lines at an angle of 45°, apply the formula for the angle between two lines.

Answer 1

The Correct Option is C

Step 1: Condition for Perpendicular Lines. For the general second-degree equation of the form: \[ Ax^2 + 2Bxy + Cy^2 = 0, \] the condition for the lines to be perpendicular is: \[ B^2 - AC = 0. \] In our case, the equation is: \[ 6x^2 - xy + \lambda y^2 = 0, \] which means:
\( A = 6 \),
\( B = -\frac{1}{2} \),
\( C = \lambda \).
Substitute these into the condition \( B^2 - AC = 0 \): \[ \left( -\frac{1}{2} \right)^2 - 6 \cdot \lambda = 0 \quad \Rightarrow \quad \frac{1}{4} - 6\lambda = 0 \quad \Rightarrow \quad \lambda = \frac{1}{24}. \] Step 2: Condition for Lines Inclined at an Angle of 45°. Using the formula for the angle between two lines: \[ \tan \theta = \frac{2\sqrt{B^2 - AC}}{A + C}, \] for \( \theta = 45^\circ \), we have: \[ 1 = \frac{2\sqrt{B^2 - AC}}{A + C}, \] leading to: \[ (A + C)^2 = 4(B^2 - AC). \] Substitute \( A = 6 \), \( B = -\frac{1}{2} \), and \( C = \lambda \): \[ (6 + \lambda)^2 = 4\left( \frac{1}{4} - 6\lambda \right), \] \[ (6 + \lambda)^2 = 4\left( \frac{1}{4} - 6\lambda \right) \quad \Rightarrow \quad 36 + 12\lambda + \lambda^2 = 1 - 24\lambda, \] \[ \lambda^2 + 36\lambda + 35 = 0. \] Solving for \( \lambda \): \[ \lambda = \frac{-36 \pm \sqrt{1296 - 140}}{2} = \frac{-36 \pm \sqrt{1156}}{2} = \frac{-36 \pm 34}{2}. \] Thus, we have two possible solutions: \[ \lambda = -1 \quad {or} \quad \lambda = -35. \] Step 3: Conclusion. Thus, the correct values of \( \lambda \) are \( -6 \) and \( -35 \).