Condition for plane $ax+by+cz+d = 0$ to touch surface $px^2+qy^2+2z=0$ Q. Show that the plane
$$ax+by+cz+d = 0$$
touches the surface
$$px^2+qy^2+2z=0$$
if $a^2/p + b^2/q +2cd = 0$.
How to start to solve this problem?

Using pole-and-polar relation

*

*Polar for a pole $(X,Y,Z)$
$$pXx+qYy+(z+Z)=0$$

*

*Identifying with the given plane:

$$\frac{pX}{a}=\frac{qY}{b}=\frac{1}{c}=\frac{Z}{d}$$

*

*For tangency, $(X,Y,Z)$ should be on the quadric, that is

\begin{align}
  0 &= pX^2+qY^2+2Z \\
  0 &=
  p\left( \frac{a}{pc} \right)^2+
  q\left( \frac{b}{qc} \right)^2+
  \frac{2d}{c} \\
  0 &= \frac{a^2}{p}+\frac{b^2}{q}+2cd
\end{align}
