Math bits

The nearest point on an ellipse to a given point

An ellipse in the coordinates orientated alone its major and minor axes is given as,

$\begin{cases} \begin{array}{c} x=a\cos t\\ y=b\sin t \end{array} & 0\le t<2\pi\end{cases}$

The squared distance from a point on ellipse to a given point(x,y),

$\begin{array}{rcl} s^{2}&=&(x-a\cos t)^{2}+(y-b\sin t)^{2}\\ &=&x^{2}+y^{2}+a^{2}\cos^{2}t+b^{2}\sin^{2}t-2xa\cos t-2yb\sin t \end{array}$

The stationary points at the zero points of its first order derivative of t,

$\frac{d(s^{2})}{dt}=-a^{2}\sin2t+b^{2}\sin2t+2xa\sin t-2yb\cos t=0$

This stationary condition is a quartic equation of cos t. With variable change u = cos t,

γ2u4 − 2αγu3 + (α2 + β2 − γ2)u2 + 2αγu − α2 = 0

where α = 2ax, β = 2by, and γ = 2(a2b2).

For all solutions from the quartic equation, the minimum distance point is identified the convex condition,

$\frac{d^{2}(s^{2})}{dt^{2}}=\gamma(1-2u^{2})+\alpha u+\frac{\beta^{2}u}{\alpha-\gamma u}>0$

$remainder(x - 0.5 \times a,a)+0.5*a$