Question

The value of 'a' for which the equation \((a^2-3)x^2 + 16xy - 2ay^2 + 4x - 8y - 2 = 0\) represents a pair of perpendicular lines is

• A. 2
• B. -1
• C. 3
• D. 4

Hint:

Always check the determinant condition \(\Delta = 0\) after finding values from the perpendicularity condition (\(A+B=0\)), as the latter only guarantees perpendicularity *if* the locus is lines.

Answer 1

The Correct Option is C

Step 1: Condition for Perpendicular Lines:
For a general second degree equation \(Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0\) to represent perpendicular lines, the sum of the coefficients of \(x^2\) and \(y^2\) must be zero. \[ A + B = 0 \] Here, \(A = a^2 - 3\) and \(B = -2a\). \[ (a^2 - 3) + (-2a) = 0 \] \[ a^2 - 2a - 3 = 0 \] \[ (a - 3)(a + 1) = 0 \] So, \(a = 3\) or \(a = -1\).
Step 2: Condition for Pair of Lines (\(\Delta = 0\)):
The equation must represent a pair of lines, so the determinant \(\Delta\) must be zero. \[ \Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \] Parameters: \(2H = 16 \implies H=8\), \(2G=4 \implies G=2\), \(2F=-8 \implies F=-4\), \(C=-2\). Check \(a=3\): \(A = 3^2-3=6\), \(B = -6\). \(\Delta = (6)(-6)(-2) + 2(-4)(2)(8) - 6(-4)^2 - (-6)(2)^2 - (-2)(8)^2\) \(= 72 - 128 - 96 + 24 + 128\) \(= 72 + 24 - 96 = 0\). So, \(a=3\) is valid. Check \(a=-1\): \(A = 1-3=-2\), \(B = 2\). \(\Delta = (-2)(2)(-2) + 2(-4)(2)(8) - (-2)(-4)^2 - (2)(2)^2 - (-2)(8)^2\) \(= 8 - 128 + 32 - 8 + 128\) \(= 32 \neq 0\). So, \(a=-1\) does not represent a pair of lines. Final Answer:
\(a = 3\).