Question

If \( 4x^2+12xy+9y^2+2gx+2fy-1=0 \) represent a pair of parallel lines then

• A. \( \frac{f}{g} + \frac{g}{f} + \frac{13}{6} = 0 \)
• B. \( f^2 + g^2 = fg \)
• C. \( f^2 + g^2 = 6fg \)
• D. \( \frac{f}{g} + \frac{g}{f} = \frac{13}{6} \)

Hint:

For parallel lines, the terms \( g \) and \( f \) are proportional to \( \sqrt{a} \) and \( \sqrt{b} \). Here \( \sqrt{4}=2, \sqrt{9}=3 \), so \( g:f = 2:3 \) or \( f:g = 3:2 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

For a general equation of the second degree to represent parallel lines, the quadratic terms must form a perfect square, and the ratio of the coefficients of linear terms must correspond to the slope defined by the quadratic part.
Step 2: Key Formula or Approach:

For \( ax^2 + 2hxy + by^2 + \dots = 0 \) to be parallel lines: \( h^2 = ab \) and \( \frac{a}{h} = \frac{h}{b} = \frac{g}{f} \).
Step 3: Detailed Explanation:

The term \( 4x^2 + 12xy + 9y^2 \) can be written as \( (2x + 3y)^2 \). This implies the lines are of the form \( 2x + 3y + c_1 = 0 \) and \( 2x + 3y + c_2 = 0 \). Their combined equation is: \[ (2x+3y)^2 + (c_1+c_2)(2x+3y) + c_1c_2 = 0 \] \[ 4x^2 + 12xy + 9y^2 + 2(c_1+c_2)x + 3(c_1+c_2)y + c_1c_2 = 0 \] Comparing with given equation \( 4x^2 + 12xy + 9y^2 + 2gx + 2fy - 1 = 0 \): Coefficient of x: \( 2(c_1+c_2) = 2g \implies c_1+c_2 = g \) Coefficient of y: \( 3(c_1+c_2) = 2f \implies c_1+c_2 = \frac{2f}{3} \) Equating the expressions for \( c_1+c_2 \): \[ g = \frac{2f}{3} \implies \frac{f}{g} = \frac{3}{2} \] We need to evaluate \( \frac{f}{g} + \frac{g}{f} \): \[ \frac{f}{g} + \frac{g}{f} = \frac{3}{2} + \frac{2}{3} = \frac{9+4}{6} = \frac{13}{6} \] This matches Option (D).
Step 4: Final Answer:

The relation is \( \frac{f}{g} + \frac{g}{f} = \frac{13}{6} \).