Normal vector to ellisoid surface

Question

How do I obtain a normal vector to the surface of an ellisoid, at a given latitude and longitude. Getting the same normal vector for a sphere is trivial.

Answer 1

An ellipsoid is given by the equation $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1, $$ for some constants $a,b,c \in \mathbb{R}_+$. If you take any curve $\alpha(t) = (x(t),y(t),z(t))$ on the ellipsoid, you know that it satisfies $$ \frac{x(t)^2}{a^2} + \frac{y(t)^2}{b^2} + \frac{z(t)^2}{c^2} = 1. $$ Differentiating with respect to $t$ gives $$ \frac{2x\dot x}{a^2} + \frac{2y \dot y}{b^2} + \frac{2z \dot z}{c^2} = 0 \iff \left\langle \left( \frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2} \right), \dot \alpha \right\rangle = 0. $$ Since $\alpha$ was arbitrary, this shows that $$(x,y,z) \mapsto\left( \frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2} \right)$$ is orthogonal to any tangent of the ellipsoid, and is therefore normal to the ellipsoid. After inserting the spherical coordinates, this becomes $$(\theta, \varphi) \mapsto\left( \frac{\cos\theta\cos\varphi}{a}, \frac{\cos\theta\sin\varphi}{b}, \frac{\sin\theta}{c} \right).$$

Answer 2

In general, consider a smooth surface $S$ defined by an equation $f(x,y,z)=0$. If $P=(x_0,y_0,z_0)$ is a point on $S$, then a normal vector at $P$ to $S$ is $$\left(\frac{\partial f}{\partial x}(x_0,y_0,z_0), \frac{\partial f}{\partial y}(x_0,y_0,z_0), \frac{\partial f}{\partial z}(x_0,y_0,z_0)\right).$$ Divide that by its length if you insist on a unit vector

In your example, the equation of an ellipse centred at the origin has $$f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1.$$ A normal vector is then $$\left(\frac{x_0}{a^2},\frac{y_0}{b^2},\frac{z_0}{c^2}\right).$$