# Minimal distance from (arbitrary) point to ellipse as the point goes to infinity

I want to solve the following problem:

Consider the ellipse $$E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$$ where $$a,b>0$$, and the point $$p(t)=(at,bt),$$ where $$t\in(0,+\infty).$$ Let $$q(t)\in E$$ be the point that minimizes the distance between $$p(t)$$ and $$E$$. Calculate: $$\lim_{t \to +\infty}q(t).$$

So, my way to think of a solution was using Lagrange multipliers in the following steps: let $$f(x,y)=\|(x,y)-p(t)\|^{2}$$ and $$g(x,y)=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}};$$ now I should find $$x,y,\lambda$$ such that $$\nabla f(x,y) =\lambda\nabla g(x,y)$$ and $$g(x,y)=1.$$

It's not that hard to write $$x$$ and $$y$$ depending on $$\lambda,$$ but as soon I plug the values of $$x$$ and $$y$$ at the last equation to find $$\lambda$$ and then get the correct $$(x,y)$$ minimizing point, I end up with a huge polynomial of $$\lambda$$ that I hardly believe I should solve.

Is that the correct step-by-step? Is there any other clever way of doing it?

Thanks on advance for the help!!!

• Seems that the point $(a,b)\in E$ will be the closest.
– mjw
Mar 18 '20 at 2:24
• @mjw I don't think so. I'm pretty sure that this would be true if and only if $t = 1$. Mar 18 '20 at 2:26
• Well okay. If $t=0$ then $\min(a,b)$ is the distance (either $(a,0)$ or $(0,b)$). On a circle it would be true. I'll rethink it. Perhaps we need to resort to the equations ...
– mjw
Mar 18 '20 at 2:29
• @mjw I think it should limit to the point at which the normal direction to the ellipse is $(a, b)$, though I need to think of a justification for this (something about strict convexity of the ellipse, maybe?). Mar 18 '20 at 2:31
• $L=(x-at)^2+(y-bt)^2 - \lambda( \frac{x^2}{a^2}+\frac{y^2}{b^2} -1)$ Taking partial derivatives with respect to $x,y,\lambda$ and setting equal to zero: $(x,y)=\left( \frac{a^3}{\sqrt{a^4+b^6}},\frac{b^3}{\sqrt{a^6+b^4}}\right)$.
– mjw
Mar 18 '20 at 2:44

## 2 Answers

$$L=(x-at)^2+(y-bt)^2-\lambda \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1 \right)$$

$$\frac{1}{2}\frac{\partial L}{\partial x}= x-at-\frac{\lambda x}{a^2}$$

$$\frac{1}{2}\frac{\partial L}{\partial y}= y-bt-\frac{\lambda y}{b^2}$$

$$\frac{\partial L}{\partial \lambda} =1-\frac{x^2}{a^2}-\frac{y^2}{b^2}$$

Setting $$\frac{\partial L}{\partial x}=\frac{\partial L}{\partial y}=0$$

We see that

$$a^2-\frac{a^3 t}{x} = b^2-\frac{b^3 t}{y}.$$

Dividing both sides by $$t$$ and letting $$t\rightarrow \infty$$:

$$\frac{a^3}{x}=\frac{b^3}{y} \textrm{ so that } x=\frac{a^3}{b^3}y.$$

Setting $$\frac{\partial{L}}{\partial \lambda}=0$$ gives us back the equation of the ellipse. Inserting $$x=\frac{a^3}{b^3}y$$ gives us $$y$$ and similarly we can solve for $$x$$:

$$(x,y)= \left( \frac{a^3}{\sqrt{a^4+b^4}} , \frac{b^3}{\sqrt{a^4+b^4}} \right).$$

So, as discussed in the comments, I believe the answer should be $$\lim_{t \to \infty} q(t) = \left(\frac{a^3}{\sqrt{a^4 + b^4}}, \frac{b^3}{\sqrt{a^4 + b^4}}\right),$$ as mjw got in his answer. I don't have a rigorous proof for this, but this is where my geometric intuition lead me. I figured that the normal direction from $$q(t)$$ out of the ellipse should limit to the direction $$(a, b)$$, i.e. parallel to the line $$p(t)$$.

Taking this reasoning for granted, we can compute the normal at an arbitrary point $$(x, y)$$ on the ellipse. We do this by computing the gradient of the function $$F(x, y) = \frac{x^2}{a^2} + \frac{y^2}{b^2}.$$ The ellipse is a level curve of this function, and the gradient points in the direction of steepest ascent, which will be perpendicular to the level surface. Thus, the normal direction from $$(x, y)$$ will be $$\nabla F(x, y) = \left(\frac{2x}{a^2}, \frac{2y}{b^2}\right).$$ Now, we want to find the $$(x, y)$$ on the ellipse such that this normal direction is parallel to $$(a, b)$$ (or equivalently, $$p(t)$$ for all $$t$$). These vectors will be parallel if and only if $$0 = \det \begin{pmatrix} \frac{2x}{a^2} & \frac{2y}{b^2} \\ a & b \end{pmatrix} = \frac{2x}{a^2} \cdot b - \frac{2y}{b^2} \cdot a.$$ Solving, we get $$y = \frac{b^3}{a^3}x.$$ Since $$(x, y)$$ lies on the ellipse, $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies \frac{x^2}{a^2} + \frac{b^4 x^2}{a^6} = 1 \implies x^2 = \frac{a^6}{a^4 + b^4}.$$ Similarly, $$y^2 = \frac{b^6}{a^4 + b^4}.$$ Clearly, out of the four possibilities for $$(x, y)$$ (including the possibilities of positive and negative coordinates), the one in the first quadrant will be closer to $$p(t)$$ than the others. So, we take the positive square roots.

Again, I have no rigorous reason to say that the limit must be the point whose normal is parallel to $$(a, b)$$. But I'm posting the answer at mjw's request anyway.

• If you take two points on the ellipse, point $A$, and point $B$, and draw segments $\overline{AP}$ and $\overline{BP}$ to some point $P$, if $\overline{AP}$ is normal to the ellipse, and $\overline{BP}$ is not, then you can, because of the convexity of the ellipse, draw a right triangle that has $\overline{AP}$ as a leg, there is a segment $\overline{BC}$ that intersects $BP$ at point $C$ with $\overline{PC}$ the hypotenuse of the triangle. Thus $AP<BP$.
– mjw
Mar 19 '20 at 15:11
• @mjw Corrected, on both counts. I do understand that a point will project onto a convex set orthogonally. It's just that I can't see a nice geometric reason why this projection must be continuous "at infinity". Think about if it were a line perpendicular to $(a, b)$ instead of an ellipse; I couldn't just choose a point indiscriminantly (with a normal parallel to $(a, b)$) and expect $q(t)$ to tend to this point. I think there needs to be something to do with the rotundity of the ellipse. Mar 21 '20 at 5:33
• Right. This rotundity property is called "curvature." mathworld.wolfram.com/Curvature.html
– mjw
Mar 22 '20 at 4:26
• Please see this posting for the curvature of an ellipse at a point $(x,y)$. math.stackexchange.com/questions/1451959/curvature-of-ellipse
– mjw
Mar 22 '20 at 4:32
• @mjw I come from a functional/convex analysis background; I was more referring to rotund norms/balls. It's a less fine tool than curvature, but I think it should do the trick, in principle. Mar 22 '20 at 5:08