How to find the minute points of an ellipse clock, knowing the minor axis and the major axis? I want to make an analogic clock, not circle, but ellipse. So the distance between minute points is not constant. I guess it grows proportionally with the division of major axis with minor axis.
How can i find these points on the ellipse ?
 A: Suppose the ellipse is not rotated, i.e.
 $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
"3" is easy. "2" should satisfy $$\frac{y}{x}=\tan\frac{2\pi}{12}=\frac{1}{\sqrt{3}}$$ or $x=\sqrt{3}y$.  Substituting we get $$\left(\frac{3}{a^2}+\frac{1}{b^2}\right)y^2=1$$
Solve for $y$, then $x=\sqrt{3}y$.

Update, per request:
  To calculate the point at 14 minutes, that is angle $\theta=\frac{2\pi}{60}$.  To calculate the point at 13 minutes, that is angle $\theta=2\frac{2\pi}{60}$.  To calculate the point at 12 minutes, that is angle $\theta=3\frac{2\pi}{60}$.  Once $\theta$ is known, start with $\frac{y}{x}=\tan\theta$, solve for $x$ (or $y$), then substitute into the ellipse equation.

A: To get the points, parametrized by time, first note that
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
and
$$
\frac{x}{y}=\tan\left(t\frac\pi{6\text{ hrs}}\right)
$$
which solved simultaneously yield the parametric curve
$$
(x,y)=\frac{ab\left(\sin\left(t\frac\pi{6\text{ hrs}}\right),\cos\left(t\frac\pi{6\text{ hrs}}\right)\right)}{\sqrt{b^2\sin^2\left(t\frac\pi{6\text{ hrs}}\right)+a^2\cos^2\left(t\frac\pi{6\text{ hrs}}\right)}}
$$
Example:
$\hspace{3.3cm}$
