Prove that the semi-vertical angle of the right circular cone $4(x^2+y^2)-9z^2=0$ is $\tan^{-1}{3/2}$.
Attempt:
we have $\cos{\alpha}=\frac{ax+by+cz}{\sqrt{x^2+y^2+z^2}\sqrt{a^2+b^2+c^2}}$ where $a,b,c $ are the drs of the axis, $\alpha$ is the semi-vertical angle of the right circular cone and $(x,y,z)$ be any point on the cone. Please help me.
Rewrite the equation: $$4(x^2+y^2)-9z^2=(2x)^2+(2y)^2-(3z)^2=0$$ The $z$-axis increases $3$ times faster for any given $z$ solution. The $x$- and $y$-axis increases twice as much for any given $x,y$ solutions. Also the semi-vertical angle is equal to $x$- or $y$-axis divided by the height of the cone, because the ratio stays the same:
$$\alpha={tan}^{-1}\left(\frac{\frac{1}{2}}{\frac{1}{3}}\right)={tan}^{-1}\left(\frac{3}{2}\right)$$
Note that in the equation $\cos{\alpha}=\frac{ax+by+cz}{\sqrt{x^2+y^2+z^2}\sqrt{a^2+b^2+c^2}}\implies$ $a=b=0, c=1$ $$\cos{\alpha}=\frac{z}{\sqrt{x^2+y^2+z^2}}$$$$ \sin{\alpha}=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$$ $$tan(\alpha)=\frac{sin\alpha}{cos\alpha}=\frac{\sqrt{x^2+y^2}}{z}$$ From the equation $4(x^2+y^2)-9z^2=0\implies \sqrt{x^2+y^2}=\frac{3}{2}z$: $$tan(\alpha)=\left(\frac{\frac{3}{2}z}{z}\right)=\frac{3}{2}\implies\alpha={tan}^{-1}\left(\frac{3}{2}\right)$$