Question

The equation $4x^2+12xy+9y^2+2gx+2fy+c=0$ will represent two real parallel straight lines if

• A. $g=4,f=9,c=0$
• B. $g=2,f=3,c=1$
• C. $g=2,f=3,c$ is any number
• D. $g=4,f=9,c>1$

Answer 1

The correct option is C

$g=2,f=3,c$ is any number

Finding the values of $g,f,c$:

Given equation is $4x^2+12xy+9y^2+2gx+2fy+c=0$.

On comparing with the general equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$, we get :

$a=4$, $2h=12$ i.e. $h=6$ and $b=9$.

We know that $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ will represent two real parallel straight lines if the following condition is satisfied,

$abc+2fgh–a{f}^2–b{g}^2–c{h}^2=0$

Substituting the values of $a=4$, $h=6$ and $b=9$.

Thus,

$\begin{array}{rcl}abc+2fgh–a{f}^2–b{g}^2–c{h}^2& =& 0\\ \left(4\right)\left(9\right)c+2fg\left(6\right)-4f^2-9g^2-c.6^2& =& 0\\ 36c+12fg-4f^2-9g^2-36c& =& 0\\ 4f^2-12fg+9g^2& =& 0\\ {\left(2f\right)}^2-2.\left(2f\right).\left(3g\right)+{\left(3g\right)}^2& =& 0\\ {\left(2f-3g\right)}^2& =& 0\\ 2f-3g& =& 0\\ 2f& =& 3g\\ \frac{f}{g}& =& \frac{3}{2}\end{array}$

Thus, we can see that $f=3,g=2$ and $c$, any number will satisfy the above condition.

Hence, option (C) is the correct answer.