Given c: y^2+z^2=1, x+z=0 : 1^° what curve is it? 2^° find…
Given $c:\ y^2+z^2=1,\ x+z=0$: $1^\circ$ what curve is it? $2^\circ$ find its projections onto the coordinate planes; $3^\circ$ find the cone with vertex $S(1,1,1)$ and directrix $c$.
Explanation of Method
A plane section of a circular cylinder, its projections, and the cone over it from a given vertex.
Step 1: $1^\circ$ An ellipse
It is a section of the circular cylinder $y^2+z^2=1$ (axis $Ox$) by a plane that is not parallel to the axis. Parametrize it as
\[ (-\sin t,\ \cos t,\ \sin t),\qquad |OP|^2=1+\sin^2t\in[1,2]. \]
It is an ellipse with center $O$. The semi-axis $\sqrt2$ runs along $\{1,0,-1\}$, with endpoints $(\mp1,0,\pm1)$. The semi-axis $1$ runs along $Oy$, with endpoints $(0,\pm1,0)$.
Step 2: $2^\circ$ Projections
| plane | eliminate | projection |
|---|---|---|
| $xOy$ | $z=-x$ | circle $x^2+y^2=1$ |
| $yOz$ | $x$ | circle $y^2+z^2=1$ |
| $xOz$ | $y$ | segment $x+z=0,\ |x|\le1$ |
Step 3: $3^\circ$ The cone
$SM$ meets the plane $x+z=0$ at $t=\tfrac{2}{2-x-z}$. The trace is
\[ Y_0=\frac{2y-x-z}{2-x-z},\qquad Z_0=\frac{z-x}{2-x-z}. \]
Requiring $Y_0^2+Z_0^2=1$: $(2y-x-z)^2+(z-x)^2=(2-x-z)^2$.
\[x^2+4y^2+z^2-4xy-4yz-2xz+4x+4z-4=0\]
Checks: $S$, and the curve points $(0,1,0)$ and $(-1,0,1)$, all satisfy it. $\checkmark$
Result
\[\boxed{\,x^2+4y^2+z^2-4xy-4yz-2xz+4x+4z-4=0\,}\]