Intersection of a cone and a plane

Question

In $\mathbb{R}^3$, given the cone $K$ and the plane $E_c$ with the equations $4x^2=y^2+z^2$ and $z=c(1-x)$.

How do I find out which different geometric objects I get for all $c\geq 0$ if I intersect both $K$ and $E_c$?

Answer 1

Expanding on the above comment, if we substitute $z = c(1-x)$ in the equation of $K$ we get $4x^2 = y^2 + c^2(1-x)^2$, i.e. $$ (4-c^2)x^2 - y^2 + 2c^2 x = c^2 \label{eq:1} \tag{1} $$ Now all we have to do is distinguish a few cases.

If $c = 0$ then \eqref{eq:1} becomes $$ 0 = 4x^2 - y^2 = (2x + y)(2x - y) $$ so $E_0 \cap K$ is a degenerate conic, the union of the lines $y = \pm 2x$.

If $0 < c < 2$ then \eqref{eq:1} becomes the equation of a hyperbola, because the terms in $x^2$ and $y^2$ have opposite signs. In particular, for $c = \sqrt{3}$ we have a rectangular hyperbola.

If $c = 2$ then \eqref{eq:1} becomes $$ y^2 = 8 x - 4 $$ so $E_2 \cap K$ is a parabola.

If $c > 2$ then \eqref{eq:1} becomes the equation of an ellipse, because the terms in $x^2$ and $y^2$ have the same sign. In particular, for $c = \sqrt{5}$ we have a circle.