Show that the semi vertical angle of the right circular cone $$4(x^2+y^2)-9z^2=0$$ is $$\arctan\left(\frac{3}{2}\right)$$
I took a cone whose vertex is at zero and axis is z axis , then I found the equation of cone and compared it with the given equation and had the answer.
Substitute $y=0$ and get $4x^2-9z^2=0$
factor $(2x-3z)(2x+3z)=0$ which gives the equation of the two lines $x=\frac{3}{2}\,z$ and $x=-\frac{3}{2}\,z$.
The semivertical angle is the angle formed by one of this lines in the plane $y=0$ with the simmetry axis of the cone. The slope $\frac{3}{2}$ is the $\tan$ of the requested angle $\alpha$, so we have $\alpha=\arctan\left(\frac{3}{2}\right)$