What should the lines be so that all segments have equal areas?

Question

I want to divide an ellipse into $n$ equal parts with vertex at one of the foci.

I chose $n=12$ and drew six lines in the ellipse:

• A line that goes through points $(0,b)$ & $(f,0)$: $y=-\frac{b}{f}x+b$, and the opposite line $y=\frac{b}{f}x-b$.
• A line that goes through $(f,0)$ & $(-f,\frac{b^2}{a})$: $y=-\frac{b^2}{2af}x+\frac{b^2}{2a}$, and the opposite line $y=\frac{b^2}{2af}x-\frac{b^2}{2a}$.
• And the lines $x=f$, & $y=0$.

I plugged $a=5$, $b=3$, and $f=4$; and the result using Desmos graphing calculator:

Clearly the elliptical segments don't have equal areas (each two segments have equal areas); per Hagen's comment that's because I have lines that create opposite segments.

What should the lines be so that all areas are the same? Can we obtain an answer regarding $n$ parts in general or we have to choose a number for $n$ like I did? Of course I need parametrical equations like what I wrote in the question body, not numerical like $y=2x+3$.

I also tried integration but it's very complicated.

Note: This question has nothing to do with Kepler's second law. Please approach it in a geometrical and algebraic sense.

Answer 1

Here's the solution.

The parametric equation of the ellipse is

$ r(t) = (a \cos t , b \sin t ) $

Now you just need to compute $t_0, t_1, t_2, \dots , t_{12} $ such that

$ A = \dfrac{1}{2} \displaystyle \bigg| \int_{t_{k-1}}^{t_k} r(t) \times r'(t) \ dt \bigg| = \dfrac{\pi}{12} $

With $t_0 = 0 $

The lines connect the focus located at $(\sqrt{a^2 - b^2}, 0) $ to $ (a \cos t_k, b \sin t_k ) $

Here's the program to generate this segmentation.

Public Sub segmenting_an_ellipse_into_12_equal_area_segments()
Dim t(12) As Double
Dim pts(10, 2) As Double
Dim r0(2), v(2, 2) As Double
a = 12
b = 7
r0(1) = 0
r0(2) = 0
v(1, 1) = a
v(2, 1) = 0
v(1, 2) = 0
v(2, 2) = b
c = Sqr(a ^ 2 - b ^ 2)
sx = 1 / a
sy = 1 / b
cx = c * sx
'find the t's
t(0) = 0
For k = 1 To 12
' equation is  1/2 integral ( t(k-1) TO t(k) ) ( (cos t - cx , sin t ) X (-sin t, cos t ) ) = 1/6 . pi/2
' so, integral ( t(k-1) TO t(k) ) ( (cos t - cx , sin t ) X (-sin t, cos t ) ) =  pi / 6
' This simplifies to  integral ( t(k-1) TO t(k) ) [ 1 - cx cos t ] dt
' = [ t(k) - t(k-1) ] - cx [ sin(t(k)) - sin(t(k-1) ] = pi / 6
' and further into
'  t(k) - cx sin(t(k)) + ( - t(k-1) + cx sin(t(k-1)) - pi/6 ) = 0
Call find_tk(k, t, cx)
Next k
k = 1
narcs = 0
ipass = 2
Call draw_ellipse(r0, v, 1)
For k1 = 0 To 12
x1 = c
y1 = 0
x2 = a * Cos(t(k1))
y2 = b * Sin(t(k1))
Call draw_line_segment(0, x1, y1, x2, y2, 46)
Next k1
Call plot_curve(0, 1, 0, 0, 0, "", "")
End Sub
Public Sub find_tk(k, t, cx)
'   t(k) - cx sin(t(k)) + ( - t(k-1) + cx sin(t(k-1)) - pi/6 ) = 0
c0 = -t(k - 1) + cx * Sin(t(k - 1)) - p / 6
' x - cx sin(x) + c0 = 0
' derivative,  1 - cx cos(x)
x = k / 6 * p
For ic = 1 To 50
y = x - cx * Sin(x) + c0
yp = 1 - cx * Cos(x)
dx = -y / yp
If Abs(y) < 0.00000001 Or Abs(dx) < 0.00000001 Then
t(k) = x
   Exit Sub
End If
x = x + dx
Next ic
MsgBox ("Newton did not converge")
End Sub