To find area of the curves that are extension of ellipse

Question

I like to draw an ellipse via 2 fixed points and a rope between the fixed points (2 focuses). I wanted to extend the idea. Point A,B,C,D are fixed points and Point E can move freely. Point E,B,C have small pulleys without friction and also their perimeters are very small. (Take zero for theoretical calculation)

If we fix a rope on Point A then it goes Point E and then Point B then C then E and finally fix again on Point D as shown figure above. If we move E while the rope stretched, we can draw a curve similar to ellipse.

We can express the equation of the closed curve as shown graph above :

$$\sqrt{(x+a)^2+y^2}+\sqrt{(x+b)^2+y^2}+\sqrt{(x-a)^2+y^2}+\sqrt{(x-b)^2+y^2}=l-2b$$

Where $l$ is lenght of the rope.

This is a symmetric curve over x and over y lines like ellipse.

My questions:

• Is there any special name of this kind of curves?
• What is the area formula of such closed curve? Is the formula similar like circle and ellipse starts with $\pi$ such as $\pi.f(a,b,l)$ ?

I tried polar coordinate transform but I could not find the area.

We can do many combinations with different number of fixed points. There is no limit of such closed curves :

Another example is:

3 fixed points (one fixed point is with a small pulley) We can get this curve if we select Point B and C in same point $P(x_1,y_1)$ in figure above.

$$\sqrt{(x+a)^2+y^2}+2\sqrt{(x-x_1)^2+(y-y_1)^2}+\sqrt{(x-a)^2+y^2}=l$$

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First of all, I focused on the simplest case (2 fixed end point, 1 fixed point with pulley on origin)

We can get this curve if we select $b=0$ in figure above.

$$\sqrt{(x+a)^2+y^2}+2\sqrt{x^2+y^2}+\sqrt{(x-a)^2+y^2}=l$$ $x=r\cos \alpha$

$y=r\sin \alpha$

$$\sqrt{r^2+a^2-2ax}+2r+\sqrt{r^2+a^2+2ax}=l$$

$$2r^2+2a^2+2\sqrt{r^2+a^2-2ax}\sqrt{r^2+a^2+2ax}=(l-2r)^2$$

$$2\sqrt{r^2+a^2-2ax}\sqrt{r^2+a^2+2ax}=2r^2-4rl+l^2-2a^2$$

$$4(r^2+a^2-2ax)(r^2+a^2+2ax)=(2r^2-4rl+l^2-2a^2)^2$$

$$4(r^2+a^2)^2-16a^2x^2=(2r^2-4rl+l^2-2a^2)^2$$

$$4(r^2+a^2)^2-16a^2r^2\cos^2 \alpha =(2r^2-4rl+l^2-2a^2)^2$$

$$4(r^2+a^2)^2 -(2r^2-4rl+l^2-2a^2)^2=16a^2r^2\cos^2 \alpha$$

$$(4r^2-4rl+l^2)(4a^2+4rl-l^2)=16a^2r^2\cos^2 \alpha$$

If we expand the terms, We will have a polynomial with degree $3$.

$r^3+(m+n\cos^2 \alpha)r^2+tr+k=0$

And my aim is to find the area of the closed curve.

$$A = 4 \int_{0}^{\pi/2} \frac{r^2}{2} d \alpha $$

I am stuck in this point because I do not see the solution easy after here. Please help me if you see how to solve the integral.

Thanks a lot for helps

Note:We can easily find the points on X axis of the curve ,$ (+\frac{l}{4},0) ; (-\frac{l}{4},0)$

And the point on Y axis of the curve , $(0,+\frac{l}{4}-\frac{a^2}{l}) ; (0,-\frac{l}{4}+\frac{a^2}{l})$

Answer 1

As mentioned in the comments, the locus of points whose sum-distance to n given points is constant is called an n-ellipse.

The image shows a 4-ellipse; without loss of generality, we can scale the figure so that B and C are at $(\pm1,0)$ (and A and D at $(\pm a,0)$), then ignore the part of the rope stretched across BC so that $l=\mathsf{AE+BE+CE+DE}$. The locus of E is the locus of intersection of two ellipses centred on the origin:

• One has foci B and C and axes $k$ and $\sqrt{k^2-1}$
• The other has foci A and D and axes $m$ and $\sqrt{m^2-a^2}$, with $2(k+m)=l$ $$\frac{x^2}{k^2}+\frac{y^2}{k^2-1}=1; \frac{x^2}{m^2}+\frac{y^2}{m^2-a^2}=1$$

After some algebraic manipulation we find $$x^2=\frac{m^2k^2(m^2-a^2-k^2+1)}{m^2-a^2k^2}$$ $$y^2=\frac{(k^2-m^2)(m^2-a^2)(k^2-1)}{m^2-a^2k^2}$$ Since manipulating all this would be too tedious, take $a=2$ and $l=8$ as an example. The expressions reduce to $$x^2=\frac{(4-k)^2k^2(13-8k)}{16-8k-3k^2}$$ $$y^2=\frac{(8k-16)(12-8k+k^2)(k^2-1)}{16-8k-3k^2}$$ Let $X=x^2$ and $Y=y^2$. The resultant of $X(k)$ and $Y(k)$, the curve's implicit equation, is quintic in X and Y: $$-6720X^5 + (-42688Y + 524160)X^4 + (-104128Y^2 + 2625248Y - 13608000)X^3 + (-123456Y^3 + 4595473Y^2 - 50494240Y + 139856640)X^2 + (-71680Y^4 + 3405920Y^3 - 55511328Y^2 + 349444224Y - 591252480)X + (-16384Y^5 + 911616Y^4 - 19293696Y^3 + 188524800Y^2 - 792115200Y + 870912000)=0$$ And here is a pretty plot of this curve, with the red dots at $(\pm1,0)$ and $(\pm2,0)$:

Now for the main part of my answer: it is not possible to find the area bounded by this curve analytically, whether by parametric equation or resultant, because the equations for x and y alone contain a square root within them that prevents the use of integration techniques for rational functions, and because the quintic expression in X and Y is not solvable. What I would do to find its area numerically is the following:

• For many positive values of x, find the corresponding y-value (which will also be positive)
• Use the points sampled to estimate the area of the curve in the first quadrant
• Since the curve is symmetric about the x- and y-axes, multiply the estimate by 4 to get the total area

Doing so with the above curve yields a result of 7.614914 (to 6 decimal places).

The computations for the 3-ellipse are similar, but involve the intersection of an ellipse with a circle.