I know that we can find the tangent of the ellipse in 2D by taking the derivative of the equation defining the ellipse.
But I'm little bit confused about finding the ellipse tangent in 3D. Where the ellipse orientation could be in any direction.
Suppose we have the following information:
- The center of an ellipse in 3D
(cx, cy, cz). - The surface normal of the ellipse
(<v1; v2; v3>). - The major and minor axis.
How can we find the tangent of ellipse at a given point on the ellipse.
Kindly help.
Best Answer
Let $\mathbf c = (cx,cy,cx)$ denote the center, and let $a$ and $b$ denote the semi-axis lengths, or "radii", as usual, and let $\mathbf u$ and $\mathbf v$ be two unit vectors in the directions of the major and minor axes respectively. Then the equation of the ellipse can be written as $$ \mathbf x(\theta) = \mathbf c + (a\cos\theta)\mathbf u + (b\sin\theta)\mathbf v $$
Then the first derivative vector is given by $$ \mathbf x'(\theta) = \mathbf c - (a\sin\theta)\mathbf u + (b\cos\theta)\mathbf v $$ This derivative vector gives you the direction of the tangent line at the point $\mathbf x(\theta)$ on the ellipse.