If $\cot x\cot y=a$ and $x+y=\frac{\pi }{6}$, then the quadratic equation satisfying cot x and cot y is

$x+y=\frac{\pi }{6}$

Apply Cot on both sides

$cot\left(x+y\right)=cot\frac{\pi }{6}$

$\frac{cotxcoty-1}{cotx+coty}=\sqrt{3}$

$\frac{a-1}{\sqrt{3}}=cotx+coty$

Quadratic equation when the roots are cot x and cot y then

$x^2-x\left(cotx+coty\right)+cotxcoty=0$

$x^2-x\left(\frac{a-1}{\sqrt{3}}\right)+a=0$

X replace by t will get

$\sqrt{3}{t}^2+(1-a)t+a\sqrt{3}=0$