Question

If the pair of straight lines \[ 9x^2+axy+4y^2+6x+by-3=0 \] represents two parallel lines, then

• A. \(a=6,\; b=2\)
• B. \(a=12,\; b=4\)
• C. \(a=3,\; b=1\)
• D. \(a=-12,\; b=4\)

Hint:

For a second degree equation to represent two parallel lines: \[ H^2=AB \] along with the determinant condition \[ \Delta=0 \] must both hold.

Answer 1

The Correct Option is B

Step 1: Compare with the general equation of pair of lines.
The general second degree equation is \[ Ax^2+2Hxy+By^2+2Gx+2Fy+C=0 \] Given, \[ 9x^2+axy+4y^2+6x+by-3=0 \] Comparing coefficients, \[ A=9,\qquad 2H=a,\qquad B=4 \] \[ 2G=6 \Rightarrow G=3 \] \[ 2F=b \Rightarrow F=\frac{b}{2} \] \[ C=-3 \]

Step 2: Condition for parallel lines.
For the equation to represent two parallel straight lines, \[ H^2=AB \] Since \[ H=\frac{a}{2}, \] we get \[ \left(\frac{a}{2}\right)^2=(9)(4) \] \[ \frac{a^2}{4}=36 \] \[ a^2=144 \] \[ a=\pm12 \]

Step 3: Use the determinant condition.
For parallel lines, the determinant condition is \[ \Delta= ABC+2FGH-AF^2-BG^2-CH^2=0 \] Substitute the values: \[ (9)(4)(-3) + 2\left(3\right)\left(\frac{b}{2}\right)\left(\frac{a}{2}\right) - 9\left(\frac{b}{2}\right)^2 - 4(3)^2 - (-3)\left(\frac{a}{2}\right)^2 =0 \] Simplifying, \[ -108+\frac{3ab}{2}-\frac{9b^2}{4}-36+\frac{3a^2}{4}=0 \] Now use \[ a^2=144 \] \[ -108+\frac{3ab}{2}-\frac{9b^2}{4}-36+108=0 \] \[ \frac{3ab}{2}-\frac{9b^2}{4}-36=0 \] Multiply by \(4\), \[ 6ab-9b^2-144=0 \] Divide by \(3\), \[ 2ab-3b^2-48=0 \]

Step 4: Check the options.
For \[ a=12,\qquad b=4, \] \[ 2(12)(4)-3(4)^2-48 \] \[ 96-48-48=0 \] Hence, the condition is satisfied.
For \[ a=-12,\qquad b=4, \] \[ 2(-12)(4)-48-48\neq0 \] Thus this is not possible.
Therefore, \[ a=12,\qquad b=4 \]

Step 5: Final conclusion.
Hence, \[ \boxed{a=12,\; b=4} \] which corresponds to option (2).