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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 $B.$

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marked as duplicate by Rahul, Quixotic, Davide Giraudo, Alexander Gruber, Hagen von Eitzen Dec 24 '12 at 12:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What have you already attempted? Hints: 1. What is the distance between two points? 2. If your point $(x_2,y_2)$ is on the ellipse, what must be true about it (e.g. It must satisfy the equation for the ellipse). – Daryl Aug 22 '12 at 22:05
I believe my answer in the follow post maybe helpful.… – matt Aug 23 '12 at 0:11
@Jin: You have not told us what are the given data about the ellipse, e.g., whether its axes are parallel to the coordinate axes or not. Maybe the ellipse is given by $5$ of its points; then the solution of your problem looks very unintuitive. – Christian Blatter Oct 22 '12 at 11:20

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.

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What if the closest point is in the bottom half of the ellipse? – Daryl Aug 22 '12 at 22:45
@Daryl how is that different? If there exists a point $(x, y)$ s.t. $(x,y)$ is on the ellipse and the distance between $(x,y)$ to $A$ is minimal, then the above formulation will find it, no? – user2468 Aug 22 '12 at 22:50
The problem is that $y=\pm \frac{b}{a}\sqrt{a^2-x^2}$. So if the point $(x_1,y_1)$ is in the bottom half of the domain ($y_1<0)$, you won't find the shortest distance as you are neglecting the bottom half of the ellipse. – Daryl Aug 22 '12 at 23:40

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 $$\frac{d L}{d x_1}=0,\,\frac{d L}{d x_2}=0,\,\frac{d L}{d \lambda}=0.$$

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You seem to have missed a rotation. Your "arbitrary" ellipses all have axes parallel to the $x$- and $y$-axes. There are five degrees of freedom in defining an ellipse. You have only given four. For each $(a:b:c:f:g:h) \in \mathbb{RP}^5$ with $h^2-ab > 0,$ there corresponds a unique ellipse given by the equation $ax^2+2hxy+2gx+by^2+2fy+c = 0.$ I agree with looking for critical points of the family of functions $\Delta : \mathbb{R}^2 \times \mathbb{S}^1 \to \mathbb{R}$ where $\Delta((x,y),p)$ is the square of the distance between the ambient point $(x,y)$ and the point $p$ on the ellipse. – Fly by Night Aug 23 '12 at 0:24
@FlybyNight: I did notice that (eventually), but the space can be simply rotated so that the axes of the ellipse and coordinate system coincide. – Daryl Aug 23 '12 at 3:01
@Daryl - I have set this up and obtained equations $2(p_x-q_x) = \lambda (2Ap_x + Bp_y + D), 2(p_y-q_y) = \lambda (Bp_x + 2Cp_y + E), Ap_x^2 + Bp_x p_y + Cp_y^2 + Dp_x + Ep_y + F = 0$ where $(p_x,p_y)$ is a point on the ellipse and $(q_x,q_y)$ is the query point. I have tried solving these using Sage and Matlab Symbolic Toolbox for $p_x$, $p_y$, and $\lambda$, and the expressions are either completely out of control or the system cannot find a solution. Is there a way to actually write the solution in closed form? – David Doria Mar 18 '13 at 19:03

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