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 $a=$

Let $y=mx$ be a line common to the given pairs of

$a{m}^2+2m+1=0$ and $m^2+2m+a=0$

$\begin{array}{l}\therefore \frac{m^2}{2(a-1)}=\frac{m}{1-a^2}=\frac{1}{2(a-1)}\\ \Rightarrow m^2=1\text{ and }m=-\frac{a+1}{2}\\ \Rightarrow (a+1{)}^2=4\Rightarrow a^2+2a-3=0\Rightarrow a=-3,1\end{array}$

But, for $a=1,$ the two pairs have both the lines common.

Hence, $a=-3.$