Answer
$\dfrac{3}{10}$
Work Step by Step
Given: $6x^2+10xy+3y^2-6y=36$
This gives:
$A=6; B=10, C=3$
Thus, $B^2-4AC=(10)^2-4(6)(3)=100-72=28 \gt 0$
This represents the equation of a hyperbola.
Now, $\cot 2 \phi=\dfrac{A-C}{B}=\dfrac{6-3}{10}=\dfrac{3}{10}$
