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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!!!

user759562
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Odylo Abdalla Costa
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  • Seems that the point $(a,b)\in E$ will be the closest. – mjw Mar 18 '20 at 02:24
  • @mjw I don't think so. I'm pretty sure that this would be true if and only if $t = 1$. – user759562 Mar 18 '20 at 02:26
  • 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 02:29
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    @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?). – user759562 Mar 18 '20 at 02:31
  • $L=(x-at)^2+(y-bt)^2 - \lambda( \frac{x^2}{a^2}+\frac{y^2}{b^2} -1)$ 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 02:44
  • @mjw That's good. I came to the same answer, from my hunch. – user759562 Mar 18 '20 at 02:49
  • But then the point $q(t)$ doesn't depend on $t$? Sorry, @mjw , could explain a little more what have you done? – Odylo Abdalla Costa Mar 18 '20 at 03:28
  • @mjw do u mean $\frac{...}{\sqrt{a^{4}+b^{4}}}$ ? otherwise it is not on the ellipse – acat3 Mar 18 '20 at 03:58
  • @RezhaAdrianTanuharja Certainly that's what I got. I didn't notice the powers of $6$ in the denominator when I read mjw's comment. Oops! Also, @ Odylo Abdalla Costa, we are taking the limit as $t \to \infty$, so the final answer should not depend on $t$. – user759562 Mar 18 '20 at 04:10
  • @mjw It should be over $\sqrt{a^4 + b^4}$, not $\sqrt{a^6 + b^6}$, otherwise it will not lie on the ellipse. The way I did it was compute $\nabla F$, where $F = \frac{x^2}{a^2} + \frac{y^2}{b^2}$, then find the points on the ellipse where this was parallel to $(a, b)$. When I solved these two equations simultaneously, I obtained the point $\left(\frac{a^3}{\sqrt{a^4 + b^4}}, \frac{b^3}{\sqrt{a^4 + b^4}}\right)$. – user759562 Mar 19 '20 at 10:04
  • Yes, you are right! Made the correction! ... $\nabla F || (a,b).$ That's a nice approach! – mjw Mar 19 '20 at 11:30
  • @user759562, please post your method as an answer! – mjw Mar 19 '20 at 11:36

2 Answers2

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

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

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

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

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 $x$:

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

mjw
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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.

user759562
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  • 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. – user759562 Mar 21 '20 at 05:33
  • Right. This rotundity property is called "curvature." https://mathworld.wolfram.com/Curvature.html – mjw Mar 22 '20 at 04:26
  • Please see this posting for the curvature of an ellipse at a point $(x,y)$. https://math.stackexchange.com/questions/1451959/curvature-of-ellipse – mjw Mar 22 '20 at 04:32
  • @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. – user759562 Mar 22 '20 at 05:08
  • please give more details. It's been a while since I've opened a book on functional analysis! In any case, I'm not an expert. – mjw Mar 22 '20 at 05:18