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Can any smooth planar curve which is closed, be a base for a 3 dimensional cone?

Lets say a vertex V is given as $(\alpha ,\beta ,\gamma )$ and the base of the cone is given as $f(x,y,z)=0;lx+my+nz=p$. How to find the equation of such a cone.

Can you give me a basic idea, a way of attacking such problems? I understand its possible there might not be an algorithm as such, but if you were to solve such a question, how would you go about it?

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marked as duplicate by J. M., William, Rudy the Reindeer, Ayman Hourieh, Douglas S. Stones Oct 6 '12 at 9:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I don't get it: is $V$ on/in the cone? How are $f(x,y,z)=0$ and $lx+my+nz$ related? – draks ... Jul 18 '12 at 9:26
V serves as the vertex of the cone, the tip of the cone if I can say so. The base of the cone is given as the curve formed when $lx+my+nz=p$ which is a plane intersects $f(x,y,z)=0$. This cross section thus formed will be the base of the cone. – Soham Jul 18 '12 at 17:12
up vote 2 down vote accepted

Assume for simplicity that $(\alpha,\beta,\gamma)=(0,0,0)$. Modulo some exceptional cases the point $(x,y,z)\ne(0,0,0)$ lies on your cone $C$ iff there is a $\lambda\ne0$ such that the point $(x',y',z'):=\bigl({x\over\lambda},{y\over\lambda},{z\over\lambda}\bigr)$ satisfies the two equations $\ell x'+ my'+ nz'=p$ and $f(x',y',z')=0$ simultaneously. From the first equation we derive that $\lambda$ would have to satisfy $\ell x+ my+ nz=p\lambda$; so necessarily $$\lambda={\ell x+ my+nz\over p}\ .$$ Plugging this into the equation $f(x',y',z')=0$ we see that $(x,y,z)\in C$ iff $$f\left({p x\over \ell x+ my+nz},{p y\over \ell x+ my+nz}, {pz\over \ell x+ my+nz}\right)=0\ .$$ The last line is the equation you were looking for.

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Thanks a lot, this is what I was looking for. Not only there is a proper explanation of how things are being approached but it cleared quite a few questions of my own. – Soham Aug 19 '12 at 16:33

I assume that $f(x,y,z)$ is a 2nd degree polynomial whose zero set intersect the plane $\ell x+my+nz=p$ in an ellipse, and that the point $V=(\alpha,\beta,\gamma)$ does not lie in the plane.

Conceptually, we just need to make some coordinate changes which simplify the picture. Let the new coordinates be denoted by $(u,v,w)$. We want the origin in the new coordinates to be the center of the ellipse mentioned above, the plane should be given by the equation $w=0$, the ellipse should be given by the equations $u^2+v^2=1$, $w=0$, and the point $V$ should have new coordinates $(u,v,w)=(0,0,1)$. Then the cone will be given by the equation $$(w-1)^2=u^2+v^2$$ Now substitute back to $(x,y,z)$-coordinates.

To carry out this program, find the center $C$ of the ellipse. Let $u$ be the vector from $C$ along the major semiaxis to one of the two points on the ellipse furthest away from $C$. Similarly, let $v$ be the vector from $C$ along the minor semiaxis to one of the two points on the ellipse closest to $C$. Let $w$ be the vector from $C$ to $V$. There are your new coordinates, and the rest is a matter of computation.

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