Equation of Sheared/Skewed Cone Let $C$ be an open ended cylinder shell running along the $z$-axis, with radius $1$ - i.e, $x^2+y^2=1$.
Let $S$ be a flat plane defined by $z=x+3$.
The cylinder $C$ and flat plane $S$ intersect to form an ellipse, $E$.
What is the equation of the cone that starts at the origin, opens up towards positive $z$, and intersects this ellipse, $E$?
I would prefer the equation be in terms of $z$, and not a parametrization.
So far, seeing as how naïve I am, I tried to use a regular cone, with $z=3\sqrt{x^2+y^2}$ coming pretty close to intersecting the ellipse $E$. I think the whole cone needs to be somehow sheared so that it goes through/intersects the ellipse $E$ - however, I have no clue whatsoever how to do this.
I tried searching for this on Google to no avail - I have no clue what such a cone would be called - I called it a sheared cone, but I'm probably wrong.
 A: Any cone (or actually symmetric pair of cones, including some degenrate ones) through the origin can be described using
$$ ax^2 + by^2 + cz^2 + dxz + eyz + fxy = 0 $$
for some parameters $a$ through $f$. Now plug $z = x+3$ into that equation. You obtain
$$ ax^2 + by^2 + cx^2 + 6cx + 9c + dx^2 + 3dx + exy + 3ey + fxy = 0 $$
Collect common monomials:
$$ (a+c+d)x^2 + by^2 + (e+f)xy + (6c+3d)x + 3ey + 9c = 0 $$
Comparing this to the equation of your cylinder, you obtain this linear system of equations:
\begin{align*}
  a+c+d &= 1 \\
  b &= 1 \\
  e+f &= 0 \\
  6c+3d &= 0 \\
  3e &= 0 \\
  9c &= -1
\end{align*}
Solving this you find the equation of your cone to be
$$ \frac89x^2 + y^2 - \frac19z^2 + \frac29xz = 0 $$
and as this is a homogenous equation, you are free to multiply it by $9$ to obtain
$$ 8x^2 + 9y^2 - z^2 + 2xz = 0 $$
If you want to, you can solve this equation for $z$, taking the positive solution only:
$$ z = x + 3\sqrt{x^2 + y^2} $$
where you see a clear relation to your naive approach. But I believe the method outlined above is more general.
