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 $m^2+2m+a=0$

$\Rightarrow \frac{m^2}{2(1-a)}=\frac{m}{a^2-1}=\frac{1}{2(1-a)}$
$\Rightarrow m^2=1$ and $m=-\frac{a+1}{2}$

$\Rightarrow {(a+1)}^2=4\Rightarrow a=1,-3$

But for $a=1$ the two pairs have both the lines common,
So $a=-3$ and slope m of the line common to both the pairs is $1$

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 line is

$(x+3y)(3x+y)=0\Rightarrow 3x^2+10xy+3y^2=0$