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

Question

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!!!

Answer 1

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

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

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

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

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 $x$:

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

Answer 2

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.