Find the conical surface with vertex S(0,0,2) and directrix…

Find the conical surface with vertex $S(0,0,2)$ and directrix $x^2+y^2=9,\ z=1$.

Explanation of Method

The conical surface with vertex above a horizontal circle.

Step 1: Trace of $SM$ in $z=1$

$SM:\ (tx,\ ty,\ 2+t(z-2))$. Setting $Z=1$ gives $t=\tfrac1{2-z}$, so the trace is $\left(\tfrac x{2-z},\ \tfrac y{2-z},\ 1\right)$. Then

\[x^2+y^2=9(2-z)^2\]

Expanded: $x^2+y^2-9z^2+36z-36=0$. This is a circular cone with vertex $S$, axis $Oz$ and $\tan\alpha=3$. Checks: $(3,0,1)$ gives $9=9$, and $S$ gives $0=0$. $\checkmark$

$S(0,0,2)$ $x^2+y^2=9,\ z=1$

The cone $x^2+y^2=9(2-z)^2$ with vertex $S$: the radius grows by $3$ for each unit of height away from $S$.

Result

\[\boxed{\,x^2+y^2=9(2-z)^2\,}\]