Closest Points On Ellipsoids

Introduction

A point outside an ellipsoid has a natural nearest point on its surface. Finding that point is not simply a matter of moving toward the ellipsoid’s center: the direction of shortest distance depends on the ellipsoid’s shape. The same problem can be more subtle when the given point is inside the ellipsoid, because there may be several equally close points on the surface.

The method is to minimize squared distance subject to the ellipsoid equation. Lagrange multipliers produce candidate points, but the multiplier equations must be handled carefully: some values make a commonly used formula undefined, and a stationary point is not automatically the nearest point. The geometry of the ellipsoid provides a useful way to check candidates.

Setting Up the Problem

Consider an ellipsoid in \(n\)-dimensional space, centered at the origin and aligned with the coordinate axes. Let its positive semiaxis lengths be \(a_1,\dots,a_n\), and let the given point be \(\mathbf{p}=(p_1,\dots,p_n)\). A point \(\mathbf{x}=(x_1,\dots,x_n)\) lies on the ellipsoid when

The squared distance from \(\mathbf{x}\) to \(\mathbf{p}\) is \(\sum_{i=1}^n(x_i-p_i)^2\). Minimizing this is equivalent to minimizing half the squared distance; the factor of one half makes the derivatives simpler.

The ellipsoid is closed and bounded, so it is compact. The squared-distance function is continuous, so it attains a minimum on the ellipsoid. Moreover, its constraint is regular everywhere on the ellipsoid: its gradient is

This gradient can vanish only when every \(x_i=0\), but the origin does not lie on the ellipsoid. Therefore, the Lagrange multiplier condition applies at every constrained minimum.

The Multiplier Equations

Minimize \(F(\mathbf{x})=\frac12\sum_{i=1}^n(x_i-p_i)^2\) subject to \(g(\mathbf{x})=0\). With the convention \(\nabla F+\lambda\nabla g=\mathbf{0}\), the coordinate equations are

Whenever \(a_i^2+\lambda\neq 0\), each equation can be rearranged to give

Substituting these coordinates into the ellipsoid equation gives a one-variable equation for the multiplier:

Solving this equation gives candidate multipliers, and the coordinate formula then gives candidate points. But the formula is valid only when none of its denominators is zero. If \(\lambda=-a_j^2\), the \(j\)-th multiplier equation instead reads \(-p_j=0\). Thus this multiplier value is possible only if \(p_j=0\); when it is possible, that equation does not determine \(x_j\). Such cases must be checked in the original equations and the ellipsoid constraint rather than discarded.

Why the Normal Direction Gives a Geometric Check

The multiplier equations say that the vector from the candidate point on the ellipsoid to the given point is a scalar multiple of the ellipsoid’s normal vector:

This normal-direction condition is necessary for a closest point, but by itself it does not say whether the point is closest or farthest. For this ellipsoid, the constraint equation gives a sharper check when \(\lambda\geq 0\).

To prove this, write \(\Delta_i=y_i-x_i\). Because both points satisfy the ellipsoid equation, subtracting \(g(\mathbf{x})=0\) from \(g(\mathbf{y})=0\) gives

Consequently, \(\nabla g(\mathbf{x})\mathbin{\cdot}(\mathbf{y}-\mathbf{x})=-\sum_i\Delta_i^2/a_i^2\). Now expand the squared distance from \(\mathbf{p}\) to \(\mathbf{y}\), using \(\mathbf{p}-\mathbf{x}=\lambda\nabla g(\mathbf{x})\):

The inequality follows because \(\lambda\geq0\) and every squared term is nonnegative. It holds for every point \(\mathbf{y}\) on the ellipsoid, proving the claim. If \(\lambda>0\), equality is possible only when every \(\Delta_i=0\), so the closest point is unique. For \(\lambda<0\), this argument does not establish a minimum; compare the candidates.

Worked Example: An Exterior Point and a Global Check

Worked Example: The Center of an Ellipse

Worked Example: An Interior Point and a Singular Multiplier

A Common Mistake: Dividing by a Zero Factor

In the interior-point example, it would be incorrect to divide the second multiplier equation \((1+\lambda)y=0\) by \(1+\lambda\) without first checking whether that factor can be zero. Doing so would force \(y=0\), leave only \((2,0)\) and \((-2,0)\), and incorrectly declare \((2,0)\) closest because its distance \(1\) is less than \(3\). The omitted multiplier \(\lambda=-1\) gives two feasible points at distance \(\sqrt{2/3}<1\).

A Reliable Procedure

For a specified axis-aligned ellipsoid and target point, the following process keeps the algebra and the geometric check distinct.

The method matters in geometry and design whenever a point must be placed on a curved boundary as close as possible to a specified location. It also illustrates a general lesson in constrained optimization: the equations locate candidates, while feasibility and comparison establish which candidate solves the original problem.