Possible Duplicate:
Calculating Distance of a Point from an Ellipse Border
Given a point $A = (x_1, y_1)$ and a $2$D ellipse, how could we find a point $B = (x_2, y_2)$ on the ellipse so that it has the shortest distance between point $A$ and $B$?
The point $A$ can be anywhere on the same plane of the ellipse. If possible, please list the final expression of the point $B.$
On
An arbitrary ellipse can be parameterised as $x=h+a\cos t,\, y=k+b\cos t$ $(0\leq t\leq 2\pi)$, which is an ellipse centred at $(h,k)$, with axis lengths $a$ and $b$.
Minimising the square of the distance is an equivalent problem to minimising the distance.
Approach 1:
The point $(x_2,y_2)$ lies on the above parametised ellipse.
Then, $$f=(x_1-h-a\cos t)^2 +(x_2-k-b\sin t)$$ is the distance to any point on the ellipse.
As $d$ is a simple function of $t$ only, the value of $t$ which is minimised when $\dfrac{d f}{d t}=0$.
Approach 2:
The point $p_1=(x_1,y_1)$ can be written as $$p_1=p_2+\lambda n$$ where $p_2=(x_2,y_2)$ is the point on the ellipse and $n$ is the vector normal to the ellipse. Using the above parameterisation for the ellipse, you obtain the system of nonlinear equations $$\begin{bmatrix}x_1\\y_1\end{bmatrix}=\begin{bmatrix}h+a\cos t\\k+b\sin t\end{bmatrix}+\lambda \begin{bmatrix}b\cos t\\a\sin t\end{bmatrix},$$ which has to be solved for $t$ and $\lambda$.
Approach 3:
A third way to formulate this problem is as a constrained optimisation problem, so that this is formulated as $$\min\limits_{x_1,x_2} f = (x_1-x_2)^2+(y_1-y_2)^2$$ subject to $$\frac{(x_2-h)^2}{a^2}+\frac{(y_2-k)^2}{b^2}-1=0.$$
This can be solved using lagrange multipliers to give $$L(x_1,x_2,\lambda)=(x_1-x_2)^2+(y_1-y_2)^2+\lambda\left(\frac{(x_2-h)^2}{a^2}+\frac{(y_2-k)^2}{b^2}-1\right).$$ The optimal solution can be found by solving the system of equations $$\frac{d L}{d x_1}=0,\,\frac{d L}{d x_2}=0,\,\frac{d L}{d \lambda}=0.$$
Not the easiest. Let the unknown point $B$ be $(x, y).$ $B$ is on the ellipse so $$(x/a)^2 + (y/b)^2 = 1,$$ i.e. $y = \frac{b}{a}\sqrt{a^2 - x^2}.$ Hence the squared distance between $A = (x_1, y_1)$ and $B$ is $$\begin{eqnarray} d & = & (x-x_1)^2 + (y -y_1)^2 \\ & = & (x-x_1)^2 + \Big(\frac{b}{a}\sqrt{a^2 - x^2} - y_1 \Big)^2. \end{eqnarray}$$ That's a function in $1$ variable, namely $x$, which you can minimize using calculus.