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 combined equation of the other two lines is given by

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

Answer 1

The correct option is B $3x^2+10xy+3y^2=0$

Let $y=mx$ be the line common to the given pairs of lines.
Then, $a{m}^2+2m+1=0\cdots \left(1\right)$
and $m^2+2m+a=0\cdots \left(2\right)$
$\left(1\right)-\left(2\right),$ we get
$(a-1)m^2+(1-a)=0$
$(a-1)(m^2-1)=0$
$\Rightarrow a=1$ or $m=\pm 1$
But for $a=1,$ the two pairs have both the lines common.
For $m=-1,$ we get $a=1$ (neglected)
$\therefore m=1$ and $a=-3$

Now, $x^2+2xy+a{y}^2=0$
$\Rightarrow$$x^2+2xy-3y^2=(x-y)(x+3y)=0$
and $a{x}^2+2xy+y^2=0$
$\Rightarrow$$-3x^2+2xy+y^2=-(x-y)(3x+y)=0$
So, the combined equation of the required lines is
$(x+3y)(3x+y)=0$
$\Rightarrow 3x^2+10xy+3y^2=0$