Find the set of points on the tangent lines drawn from…

Find the set of points on the tangent lines drawn from $S(0,0,-1)$ to the surface $z=\tfrac12(x^2+y^2)$. Sketch it.

Step 1: Points of tangency

Let $P(a,b,c)$ be on the paraboloid, so $a^2+b^2=2c$. Its tangent plane is

\[ a(x-a)+b(y-b)-(z-c)=0\iff ax+by-z=c . \]

A tangent line from $S$ touches at $P$ only if $S$ lies in the tangent plane at $P$: $0+0+1=c$. So the points of tangency form the circle

\[ c=1,\qquad a^2+b^2=2 . \]

Step 2: The locus

The line $SP$ is $(ta,\ tb,\ -1+2t)$. Then $t=\tfrac{z+1}2$, $a=\tfrac{2x}{z+1}$ and $b=\tfrac{2y}{z+1}$. Using $a^2+b^2=2$: $x^2+y^2=\tfrac12(z+1)^2$.

This is a circular cone with vertex $S$, axis $Oz$ and $\tan\alpha=\tfrac1{\sqrt2}$.

Check in the plane $y=0$: the lines $z=-1\pm\sqrt2\,x$ have slope $\sqrt2$ at $x=\sqrt2$, the same as the parabola $z=\tfrac{x^2}2$, and the same height $1$. $\checkmark$

So the cone does not just meet the paraboloid along the circle $x^2+y^2=2,\ z=1$. It is tangent to it there, with the same tangent planes along the whole circle.