Question

If the slope of one of the lines in the pair of lines \( 8x^2 + axy + y^2 = 0 \) is thrice the slope of the second line, then \( a = \) ?

• A. \( 8\sqrt{\frac{2}{3}} \)
• B. \( 6 \)
• C. \( 16\sqrt{2} \)
• D. \( 3\sqrt{\frac{2}{5}} \)

Hint:

For a pair of straight lines given by \( Ax^2 + 2Hxy + By^2 = 0 \), the slopes are given by: \[ m = \frac{-H \pm \sqrt{H^2 - AB}}{B}. \]

Answer 1

The Correct Option is A

Step 1: General Equation for a Pair of Straight Lines A second-degree equation of the form:
\[ Ax^2 + 2Hxy + By^2 = 0 \] represents two straight lines passing through the origin. The slopes of these lines are given by solving: \[ m = \frac{- (H \pm \sqrt{H^2 - AB})}{B}. \] Comparing with the given equation: \[ 8x^2 + axy + y^2 = 0, \] we have: \[ A = 8, \quad 2H = a, \quad B = 1. \] Thus, \( H = \frac{a}{2} \). 

Step 2: Solve for Slopes 
The slopes of the lines are given by: \[ m = \frac{- H \pm \sqrt{H^2 - AB}}{B}. \] Substituting \( A = 8, B = 1, H = \frac{a}{2} \): \[ m = \frac{-\frac{a}{2} \pm \sqrt{\left(\frac{a}{2}\right)^2 - (8)(1)}}{1}. \] \[ m = \frac{-\frac{a}{2} \pm \sqrt{\frac{a^2}{4} - 8}}{1}. \] 

Step 3: Given Condition 
One slope is three times the other: \[ m_1 = 3m_2. \]  Cross multiplying: \[ - \frac{a}{2} + \sqrt{\frac{a^2}{4} - 8} = 3 \left(- \frac{a}{2} - \sqrt{\frac{a^2}{4} - 8} \right). \] Expanding: \[ - \frac{a}{2} + \sqrt{\frac{a^2}{4} - 8} = - \frac{3a}{2} - 3\sqrt{\frac{a^2}{4} - 8}. \] Rearrange: \[ \sqrt{\frac{a^2}{4} - 8} + 3\sqrt{\frac{a^2}{4} - 8} = -\frac{3a}{2} + \frac{a}{2}. \] \[ 4\sqrt{\frac{a^2}{4} - 8} = -a. \] \[ \sqrt{\frac{a^2}{4} - 8} = -\frac{a}{4}. \] Squaring both sides: \[ \frac{a^2}{4} - 8 = \frac{a^2}{16}. \] Multiply by 16: \[ 4a^2 - 128 = a^2. \] \[ 3a^2 = 128. \] \[ a^2 = \frac{128}{3}. \] \[ a = 8\sqrt{\frac{2}{3}}. \] 

Step 4: Conclusion 
Thus, the correct answer is \( \mathbf{8\sqrt{\frac{2}{3}}} \).