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Oblique meaning not a regular cone; its apex makes an angle to the base.

An alternate way of visualizing this: consider a sphere centered at the origin. The sphere traces a circle on the XY-plane. Now consider a point A anywhere on the sphere, and another point B on the circle. Now as B traces around the circle, length AB varies. I an trying to draw the solid (or surface) with AB as the generator.

The pics and video below would give you the final idea of the solid I want to generate:

orthographic view with the sphere View from the top without the sphere for clarity0

https://youtu.be/Z8RAn6Dkz7w - animation

https://ggbm.at/vrEC7tDs - geogebra worksheet (3d)

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The general equation for a upright cone with the tip at $Z=0$ is

$$ 0 = X^2 + Y^2 - Z^2$$

(Ignoring the points for which $Z>0$). If you consider slices for some constant $Z$ you get circles centered on the line $X=Y=0$. To make an oblique cone with circles along the line $X=aZ,Y=bZ$ we can just shift the coordinates accordingly and get the equation

$$ 0 = (X-aZ)^2 + (Y-bZ)^2 - Z^2$$

(again for $Z<0$). You can make it narrower or wider by scaling the $Z$ coordinate. To position the tip at some other point $(x,y,z)$ you can just shift the whole thing using the transform $(X',Y',Z') := (X-x,Y-y,Z-z)$.

With the same idea you can easily parametrize the (upright) cone by

$$(u,v) \mapsto (u\cos(v),u\sin(v),u)$$

or the oblique cone by

$$(u,v) \mapsto (u\cos(v)-au,u\sin(v)-bu,u)$$

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Suppose you have a circle in 3D space. It can be described by its radius $r$, its center point $\mathbf{C}$, and two orthogonal unit vectors $\mathbf{A}$ and $\mathbf{B}$ that span the plane parallel to the circle. Then the parametric equations of the circle are: $$ \mathbf{P}(u) = \mathbf{C} + r\cos(u)\mathbf{A} + r\sin(u)\mathbf{B} $$ Now suppose you want to form an oblique cone that contains this circle and an apex point $\mathbf{Q}$. The parametric equations of this surface are: $$ \mathbf{S}(u,v) = (1-v)\mathbf{Q} + v\mathbf{P}(u) $$ If $v=0$, then $\mathbf{S}(u,v) = \mathbf{Q}$ for all $u$, and if $v=0$, then $\mathbf{S}(u,v) = \mathbf{P}(u)$ is a point on the circle.

If you fix $u$, then, as $v$ varies, you travel along a straight line through the points $\mathbf{Q}$ and $\mathbf{P}(u)$.

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