# Parametric equation of a cone.

I have a cone with vertex (a, b, c) and base circumference with center $(x_0,y_0)$ and radius R. I can't understand what is the parametric representation of three dimensional space inside the cone. Any suggestions please?

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The parametric equation of the circle is:

$$\gamma(u) = (x_0 + R\cos u, y_0 + R\sin u, 0)$$

Each point on the cone lies on a line that passes through $p(a, b, c)$ and a point on the circle. Therefore, the direction vector of such a line is:

$$\gamma(u) - p = (x_0 + R\cos u, y_0 + R\sin u, 0) - (a, b, c) = (x_0 - a + R\cos u, y_0 - b + R\sin u, -c)$$

And since the line passes through $p(a, b, c)$, the parametric equation of the line is:

$$p + v\left(\gamma(u) - p\right) = (a, b, c) + v \left((x_0 - a + R\cos u, y_0 - b + R\sin u, -c)\right)$$

Simplify to get the parametric equation of the cone:

$$\sigma(u, v) = \left(a(1-v) + v(x_0 + R\cos u), b(1-v) + v(y_0 + R\sin u), c(1 - v)\right)$$

Here is a plot of the cone for $p(0, 0, 2)$, $(x_0, y_0) = (-1, -1)$ and $R = 1$, when $u$ scans $[0, 2\pi]$ and $v$ scans $[0, 1]$:

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Begin with a parametric representation of the base, using polar coordinates (with the center shifted to $(x_0,y_0)$. For any given point inside the cone, draw a straight line from the vertex through the point to the base of the cone. Let two parameters specify the latter point. Use $z$ as the third parameter, or perhaps a variable $t$ that varies linearly with $z$ so that $t=0$ at the vertex and $t=1$ at the base of the cone. Many minor variations over this theme are possible.