Question

Find the equation of the right circular cone vwhen the straight line 2y+3z=6, x=0 revolves about the z-axis

Answer 1

Given the vertex point $p_0=(x_0,y_0,z_0)=(0,0,2),$ and $p=(x,y,z)$

the support circle in the $xy$ plane is $x^2+y^2=r^2$ with $r=3.$

The generatrix line is given as $\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}$ so its intersection with the $xy$  plane is given by

$(x_0-z_0\frac{l}{n},y_0-z_0\frac{m}{n},0)$

so this point will lie on the given circle $(x_0-z_0\frac{l}{n})^2+(y_0-z_0\frac{m}{n})^2=r^2$.

Here we have $(2\frac{l}{n})^2+(2\frac{m}{n})^2=9$

but $\begin{cases}l=\frac{x}{k}\\m=\frac{y}{k}\\n=\frac{z-2}{k} \end{cases}$ substituting

$(2\frac{x}{z-2})^2+(2\frac{y}{z-2})^2=9$

and finally

$\frac{x^2}{9}+\frac{y^2}{9}-\frac{(z-2)^2}{4}=0$.