Calculating the distance between an ellipse and a point.

Question

I have the following parametrization of a curve (an ellipse): \begin{align} x(t) &=\frac{-17(\cos(t)-\sqrt{2}\sin(t))+23}{6} \\ y(t) &=\frac{17(\cos{(t)}-\sqrt{2}\sin(t))+23}{6} \\ z(t) &=\frac{17\sqrt{2}\sin(t)+26}{6} \end{align} for $0\leq t\leq2\pi$.

Now how can I calculate the distance from the point $(1,1,s)$ to this parametric curve in function only of $s$?

Answer 1

This is a nice problem, and I do wonder how it was posed to you. It seemed forbiddingly difficult, till I realized that the geometric situation made the solution considerably easier, even if ultimately impossible from any practical or computational standpoint.

Let me reveal the end of the story first: it seems clear that the answer to the question involves finding the roots of a quartic polynomial that sometimes has four real roots (depending on $s$), but usually only two. No matter how you look at it, this quartic will ordinarily be out of reach. Now, you recognize, I’m sure, that any minimization problem consists of two steps: finding the condition for minimum and then calculating its value. In our situation, once given the value of $s$, we have to find the value of $t$ that minimizes the distance between the $t$-point on the ellipse and the point $(1,1,s)$ on the line. That’s the big deal here, and you see that once you know this value of $t$, the distance is easy to find. I’ll deal with this difficult problem only.

The geometric situation is this: your ellipse $\mathcal E$ sits in the vertical plane $x+y=23/3$, and your moving point travels along the vertical line $\mathcal L$: $x=1,y=1,z=s$. Thus the line and the plane are parallel. Furthermore, the plane containing the $z$ axis and $\mathcal L$ hits the ellipse’s plane orthogonally. A really nice and simple situation! Note also that according to your formulas, the center of the ellipse is at $(23/6,23/6,13/3)$. This prompts me to take a new coordinate system $(\xi,\eta,\zeta)$ that’s rotated through the $z$-axis, $45^\circ$ clockwise from the standard one. This will put the ellipse’s plane parallel to the $\xi\zeta$-plane with the equation $\eta=23\sqrt2/6$. And the center of the ellipse will be at $(0,23\sqrt2/6,13/3$. This rotation is not necessary to the solution of the problem, but it makes it easier for me to see what’s going on.

Specifically, the transformation I’ve used is \begin{align} \xi&=\frac1{\sqrt2}(x-y)\\ \eta&=\frac1{\sqrt2}(x+y)\\ \zeta&=z\,, \end{align} good and orthogonal, so that all distances are preserved. I will work with this to describe the point $E_t\in\mathcal E$ and the point $P_s\in\mathcal L$. The formulas turn out that when we call $P_s$ the $s$-point on $\mathcal L$, \begin{align} E_t&=\,\bigr(\xi(t),\eta(t),\zeta(t)\bigr)\>,\text{where}\\ \xi(t)&=\frac{17}6(2\sin t+\sqrt2\cos t)\\ \eta(t)&=\frac{23\sqrt2}6\\ \zeta(t)&=\frac{13}3+\frac{17\sqrt2}6\sin t\\ P_s&=(0,\sqrt2,s)\,. \end{align}

Now comes the significant simplification: For a point $P$ on our vertical line, the distance to any point $Q$ in the ellipse’s plane will be given as the length of the hypotenuse of the triangle whose base is in the plane, and whose altitude is the length of the perpendicular dropped from $P$ to the plane, of length equal to the distance $\delta$ between $\mathcal L$ and the plane. It happens that $\delta=17\sqrt2/6$, but I don’t think the coincidence is significant here. At any rate, to minimize $(\delta^2+(\text{base length})^2)^{1/2}$, it is quite enough to minimize that base length. Since the perpendicular from $P_s$ to the plane hits the ellipse plane at $(0,23\sqrt2/6,s)$, we might as well ignore the $\eta$-coordinate, in other words just project to the $(\xi,\zeta)$-plane. Here’s the picture, courtesy of Desmos, with $\mathcal E$ in red:

I’ve drawn in what seem to be the minimum distance from $(0,s)$ to $\mathcal E$ as well as the maximum. If $(0,s)$ had been inside, there presumably would be two minima, two maxima. At a minimum, the tangent vector to $\mathcal E$, namely $\bigl(\xi'(t),\zeta'(t)\bigr)$, will be perpendicular to the vector from $\bigl(\xi(t),\eta(t)\bigr)$ to $(0,s)$. Since the derivative has common factor $17/6$, I’ve divided by that. Then the dot product is $$ (\sqrt2\sin t + 2\cos t)\cdot\left(\frac{17}3\sin t - \frac{17\sqrt2}6\cos t, \frac{13}3 - s + \frac{17\sqrt2}6\sin t\right)\,, $$ which works out to $$ 0=\left[-\frac{17\sqrt2}3 + \frac{34\sqrt2}3\sin^2t\right]\>+\>\left[\left(\frac{13\sqrt2}3 - \sqrt2s + \frac{34}3\sin t\right)\cos t \right]\,, $$ for perpendicularity, and thus location of the minimum (or maximum). You notice, if you care to, that the left-hand bracket is $-\frac{17\sqrt2}3\cos(2t)$, while the right-hand bracket is $\sqrt2\bigl(\frac{13}3-s\bigr)\cos t+\frac{17}3\sin(2t)$. As a check, in the special circumstance that $s=13/3$, so in the center of the ellipse, the requirement that the dot product be zero becomes $\tan(2t)=\sqrt2$, giving you the two maxima (at the vertices of the ellipse) and the two minima (at the ends of the minor axis). Geometrically, we knew that already, but it suggests that my computations may be correct.

In general, however, you probably don’t want to notice that, but rather replace $\cos t$ by $(1-\sin^2 t)^{1/2}$ and do the necessary thing to get, I’m sure, a quartic in $\sin t$, surely infeasible to find the roots of.