Question

If the pair of lines $x^2+2xy+a{y}^2=0$ and $a{x}^2+2xy+y^2=0$ have exactly one line in common, then prove that joint equation of the other two lines is given by $3x^2+10xy+3y^2=0$.

Answer 1

Let $y=mx$ be a line common to both
$\therefore a{m}^2+2m+1=0$ and $m^2+2m+a=0$
$\therefore \frac{m^2}{2(1-a)}=\frac{m}{a^2-1}=\frac{1}{2(1-a)}$
$\therefore m^2=1$ and $m=-\frac{1}{2}(a+1)$
$\therefore \frac{1}{4}(a+1{)}^2=m^2=1$
or $(a+1{)}^2-(2{)}^2=0$
or $(a+3)(a-1)=0\therefore a=-3,1$
For $a=1$, the equations become identical so that both
the equations are common. Since only one equation is
common, we choose $a=-3$.
The two pairs are $x^2+2xy-3y^2=0$
and $-3x^2+2xy+y^2=0$
or $(x+3y)(x-y)=0$ and $-(x-y)(3x+y)=0$
$x-y=0$ is the common line and the other lines are given
by $x+3y=0,3x+y=0$.
Their joint equation is $(x+3y)(3x+y)=0$
or $3x^2+10xy+3y^2=0$