Question

If the pairs of lines $x^2+2xy+a{y}^2=0$ and $a{x}^2+2xy+y^2=0$ have exactly one line in common then the joint equation of the other two lines is given by

• A. $3x^2+8xy-3y^2=0$
• B. $3x^2+10xy+3y^2=0$
• C. $y^2+2xy-3x^2=0$
• D. $x^2+2xy-3y^2=0$

Answer 1

The correct option is B

$3x^2+10xy+3y^2=0$

Let y = mx be a line common to the given pairs of lines, then
$a{m}^2+2m+1=0$ and $m^2+2m+a=0$. Solving these two equations we get,
a = 1 or -3
But for a = 1, the two pairs have both the lines common, so a =-3 and $m=-\left(\frac{a+1}{2}\right)$
$\therefore$ m = 1
Now, $x^2+2xy+a{y}^2=x^2+2xy-3y^2$
= (x - y) (x + 3y)
and $a{x}^2+2xy+y^2=-3x^2+2xy+y^2$
= -(x - y) (3x + y)
So, the equation of the required lines is
(x + 3y) (3x + y) = 0 $\Rightarrow 3x^2+10xy+3y^2=0$