Finding an ellipse knowing two points and the arc length

Question

I have two (Cartesian) points of an elipse, and I know the arc length between them, but I don't know either radii or where the centre is.

I know that one known point lies on the minor radius though.

Is it possible to get the equation at all?

If it helps: Point 1 [on the minor axis]: (0,100); Point 2: (145,725); Arc length: 650

Edit: The radii are parallel to the axes

Answer 1

First attempt: Going from a sketch in GeoGebra

the equation is roughly $$ \left(\frac{x - 513 }{513}\right)^2 + \left(\frac{y- 100}{877.5}\right)^2 = 1 $$

Algebraic model: Assuming the minor axis is parallel to the $x$-axis, the equation of the ellipse is $$ \left(\frac{x - b}{b}\right)^2 + \left(\frac{y - 100}{a}\right)^2 = 1 \quad (1) $$ with unknowns $a, b$.

Note: Other orientations parallel to one of the coordinate system axes are hardly possible, because the distance between $P$ and the second point $Q$ is $d = 641.6$, while the intended arc length between $P$ and $Q$ is $s = 650$, less then $10$ units more. I tried an orientation parallel to the $y$-axis, but that would need a much larger piece of arc on the ellipse.

Inserting the first point $P = (0, 100)$: $$ \left(\frac{0 - b}{b}\right)^2 + \left(\frac{100 - 100}{a}\right)^2 = 1 $$ It satisfies equation $(1)$, so that curve goes through $P$.

The second point $Q = (145,725)$ gives the equation $$ 1 = \left(\frac{145-b}{b}\right)^2 + \left(\frac{725 - 100}{a}\right)^2 = \left(1 - \frac{145}{b}\right)^2 + \left(\frac{625}{a}\right)^2 $$ which relates $a$ and $b$ and solving gives: $$ a = \frac{625}{\sqrt{1 - \left(1 - \frac{145}{b}\right)^2}} $$ and $$ \left(1 - \frac{625^2}{a^2}\right) b^2 = (b-145)^2 = b^2 -290 b + 145^2 \iff \\ b^2 = \frac{290}{625^2}a^2 b - \frac{145^2}{625^2}a^2 \iff \\ \left(b - \frac{145}{625^2}a^2 \right)^2 = \frac{145^2}{625^4}a^4 - \frac{145^2}{625^2}a^2 = \frac{145^2}{625^4}a^2 \left(a^2 - 625^2\right) \iff \\ b = \frac{145}{625^2}a^2 + \frac{145}{625^2} a \sqrt{a^2 - 625^2} \iff \\ b = \frac{145}{625^2}a \left(a + \sqrt{a^2 - 625^2} \right) \quad (2) $$

We choose this parameterization of the arc:

$$ x = b - b \cos t \quad y = 100 + a \sin t \\ \dot{x} = b \sin t \quad \dot{y} = a \cos t $$

Note: While $t$ runs from $0$ to $2\pi$, $t$ in general is not the angle $\beta = \angle(R, C, P)$, where $R = (x(t),y(t))$ and $C = (b, 100)$ is the center of the ellipse. For a circle ($a = b$) it would be, but for a non-circle ellipse the relationship is non-linear: $$ \beta = \arctan\left(\frac{a \sin t}{b \cos t}\right) = \arctan\left(\frac{a}{b} \tan t\right) \quad t \in [0, \pi/2] $$

The chosen curve leads to the arc length via $ds^2 = dx^2 + dy^2$: \begin{align} s &= 650 \\ &= \int\limits_{0}^{t^*} \sqrt{b^2 \sin^2 t + a^2 \cos^2 t} \, dt \\ &= a \int\limits_{0}^{t^*} \sqrt{1 - (1- (b/a)^2) \sin^2 t} \, dt \\ &= a \int\limits_{0}^{t^*} \sqrt{1 - \epsilon^2 \sin^2 t} \, dt \\ &= a \, E(t^*, \epsilon) \quad (3) \end{align} with $t^*$ determined by $$ 145 = b - b \cos t^* \quad 725 = 100 + a \sin t^* $$ where we use $$ t^* = \arcsin \frac{625}{a} \quad (4) $$ The above integral is an elliptical integral (Legendre, 2nd kind) which can only be approximated by elementary functions.

For the eccentricity $\epsilon$ we get $$ 1 - \epsilon^2 = \frac{b^2}{a^2} = \frac{145^2}{625^4} \left(a + \sqrt{a^2 - 625^2} \right)^2 \iff \\ \epsilon = \sqrt{1-\frac{145^2}{625^4} \left(a + \sqrt{a^2 - 625^2} \right)^2} \quad (5) $$

Using the above equations $(1)-(5)$ we can formulate an equation in terms of the unknown $a$: $$ 650 = a E\left(\arcsin \frac{625}{a}, \epsilon(a)\right) \iff \\ F(a) = 650 - a E\left(\arcsin \frac{625}{a}, \epsilon(a)\right) = 0 \quad (6) $$

Solving this numerically (details below), we get this solution: $$ a = 721.384624 \quad b = 289.634475 $$

and the equation $$ \left(\frac{x - 289.634475}{289.634475}\right)^2 + \left(\frac{y- 100}{721.384624}\right)^2 = 1 $$

which is quite different from the attempt to derive these values by the sketch!

Numerical solution:

Looking through my mostly open source toolset I decided to use Maxima, because it features an implementation of the needed incomplete elliptic integral of the second kind:

First we define $t^*(a)$:

(%i) ts(a) := asin(625/a);
                                         625
(%o)                       ts(a) := asin(---)
                                          a

Then we define $m(a) = \epsilon^2(a)$, the second needed parameter for the elliptic integral implementation elliptic_e(phi, m):

(%i) m(a) := 1-((145^2)/(625^4))*(a + sqrt(a^2 - 625^2))^2;

                                2
                             145             2      2  2
(%o)             m(a) := 1 - ---- (a + sqrt(a  - 625 ))
                                4
                             625

And then we define $F(a)$:

(%i) F(a):= 650 - a*elliptic_e(ts(a), m(a));
(%o)             F(a) := 650 - a elliptic_e(ts(a), m(a))

Now we need a way to find roots $a$ of $F(a)$. First we try it graphically:

(%i) plot2d(F(a), [a,625,1000]);
(%o) 

So the root is around $725$. Next we try the Newton method:

(%i) load(newton1);
(%o)       /usr/share/maxima/5.27.0/share/numeric/newton1.mac

then we execute it using $a_0 = 800$ and a precision $10^{-6}$:

(%i) an : newton(F(a), a, 800, 1/100000);
(%o)                        721.3846241818675

The result is saved as $a_n$. From this we calculate $b_n$:

(%i) bn = sqrt(1-m(an)) * an;
(%o)              289.6344752045375 = 289.6344752045375

For testing if these values fulfill equation $(1)$ we define

(%i) eq(a,b) := ((145-b)/b)^2 + ((725-100)/a)^2;
                               145 - b 2    725 - 100 2
(%o)              eq(a, b) := (-------)  + (---------)
                                  b             a

and apply this test function to the calculated values

(%i) eq(an, bn);
(%o)                               1.0

And this looks good! Some more tests:

(%i) eq(x,y,a,b) := ((x-b)/b)^2 + ((y-100)/a)^2;
                                    x - b 2    y - 100 2
(%o)             eq(x, y, a, b) := (-----)  + (-------)
                                      b           a
(%i) eq(0,100,an,bn);
(%o)                               1.0
(%i) eq(145,725,an,bn);
(%o)                               1.0
(%i) eq(513,877,an,bn);
(%o)                        1.754880658615442
(%i) eq(2*bn,100,an,bn);
(%o)                               1.0
(%i) eq(bn,100-an,an,bn);
(%o)                               1.0
(%i) eq(bn,100+an,an,bn);
(%o)                               1.0

And finally some looks at the arc length:

(%i) F(an);
(%o)                     - 6.5450649344711564E-9
(%i) ts(an);
(%o)                        1.047926026242998
(%i) an * elliptic_e(ts(an), m(an)),numer;
(%o)                        650.0000000065451
(%i) an * elliptic_e(%pi/2, m(an)),numer;
(%o)                        830.6877077032057
(%i) an * elliptic_e(0, m(an)),numer;
(%o)                               0.0

Phew.

Answer 2

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I offer the above picture as a proof without words to show that, even if the afore-mentioned ellipse were uniquely determined in terms of size and dimensions $($and I'm not saying that it is$)$, it cannot be uniquely determined in terms of position. Ellipse #$2$ is the reflection of Ellipse #$1$ with regard to $AB$, whereas Ellipse #$3$ and Ellipse #$4$ are the reflections of Ellipse #$1$ and Ellipse #$2$, respectively, with regard to the mediating bisector of the line segment $(AB)$.