Parametric Equations of an Oblique Circular Cone

Question

I am trying to determine the parametric equations for a specific shape of an oblique circular cone with no success. Exhaustive web searchs and many texts have not been fruitful as regards paramaterization of oblique circular cones. Since I do not know of a name for the shape of the cone needed, I will describe its properties. For reference, the orientation of the cone is with the apex at point (0, 0, 0) and opens along the +z axis to height "h" with the base diameter "b". The circular sections in the x-y plane are constrained to be such that $0 \leq y \leq b$ (that is, y values are always positve in quadrants I and II of the x-y plane) and $-\frac b2 \leq x \leq \frac b2$. The axis is the line through the points (0, 0, 0) and (0, $\frac b2$, h). The point (0, 0) of each circle in the x-y plane must be at the point z, that is, (0, 0, z). The shape of the cone then will be a right triangle in 2D when view from the x-axis with the cone height being a vertial line along the z-axis and the longest slant being from (0, 0, 0) to (0, b, h). A crude 2D text picture of the desired cone shape is below with the z-axis horizontal (+z to the left) and the y-axix vertical (+y downward). Can anyone help with the parameterization of such a shaped cone? Obviously, I have not worked out the implicit equation either although it seems it should simply be some variation of the form $x^2 + y^2 = z^2$. (The cone orientation described here is for simple reference only and ideally the parameterization should work in any orientation.) Thanks!

    + z ----------------------
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Answer 1

Suppose we have a curve $C(u)$ and a point $P$, and we want a parametric equation for the cone that has its apex at $P$ and contains the curve $C$. A suitable equation is $$ S(u,v) = (1-v)P + vC(u) $$ You can see that $S(u,0) = P$ and $S(u,1) = C(u)$ for all $u$. Also, if we fix $u$, then the curve $v \mapsto S(u,v)$ is a straight line passing through the points $P$ and $C(u)$.

In your case, you can use any of the circles parallel to the $xy$ plane for $C$. For example, you could use the circle in the plane $z=h$, which is $$ C(u) = \left(0, \tfrac{b}2, h\right) + \left(\tfrac{b}2 \cos u, \tfrac{b}2 \sin u, 0\right) \quad (0 \le u \le 2\pi) $$ I checked the case $b=4$, $h=7$. Here is the Mathematica code:

P = {0, 0, 0};
Cu = {0, b/2, h} + {(b/2)*Cos[u], (b/2)*Sin[u], 0};
Suv = (1 - v)*P + v*Cu;
fix = {b -> 4, h -> 7};
f = Suv /. fix;
ParametricPlot3D[f, {u, 0, 2*Pi}, {v, 0, 1}]

and here is the image it produced: