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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.

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up vote 2 down vote accepted

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.

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Tyvm - that worked; and I like your general formula. I should have been able to figure this out on my own tbh.. >_> – Shariq Dec 10 '12 at 4:01
I have one solution that is so easy. If someone wants, he/she call me. – bigli Mar 12 '15 at 13:40
@bigli: Why not post your solution as an alternative answer? – MvG Mar 13 '15 at 17:29

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