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A cone in 3 dimensions has a vertex and a base. The contour of the base is a circle which is a smooth closed planar curve. Can there be a more general cone which can have any smooth closed planar curve as the contour of its base ?

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up vote 3 down vote accepted

Certainly; if you have some plane curve represented parametrically as $(f(u)\quad g(u))^T$, the surface represented parametrically by

$$\begin{align*}x&=f(u)(1-v)\\y&=g(u)(1-v)\\z&=v\end{align*}$$

is one simple parametrization for a generalized (right) cone whose cross sections are scaled versions of the given plane curve.

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M.: Thank you for the answer. –  Rajesh D Apr 24 '11 at 13:29
    
Of course, if what you need is a skew cone (axis not perpendicular to a coordinate plane), a different approach is needed... –  J. M. Apr 24 '11 at 13:32
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