Region bounded by Ellipse and Line

Question

I have an ellipse defined either as algebraic or geometric:

algebraic: ax^2 + bxy +cy^2 + dx +ey +f = 0

geometric: major Axis, minor Axis, x center, y center, rotation angle from horizontal

And I have a line that could be defined by two points (x1,y1) and (x2,y2)

These two objects would intersect each other:

Question

How could I model the two objects, so that:

For an ELLIPSE, I could get -

1) Length of the line within the ellipse

2) Area of the yellow shaded region (i.e. smaller area)

Attempt

Previously I was able to get the area for a rotated ellipsoid under a horizontal plane. But now I would like to arbitrarily model that line (instead of a plane).

Thanks for your help!

Regards Caid

*UPDATE:

For a 3D ellipsoid:

`Nvolume[p_, abc_] :=

I was able to obtain the volume bounded under a plane defined as: p = {Sin[k Degree], 0, Cos[k Degree]};

And then, the planar area:

Unfortunately, that's about what I have. I need to convert the plane into a line for a 2D case.

Also if possible, I would like to obtain general equations for both.

Answer 1

I would suggest to use the parametric form of a centered, rotated ellipse

{a Cos[t] Cos[Theta] - b Sin[t] Sin[Theta], a Cos[t] Sin[Theta] + b Sin[t] Cos[Theta]}

with a,b as the semimajor and semiminor axis, Theta as the rotation angle of the ellipse (0 for an axis oriented ellipse).

The trick is to rotate and center your problem into a configuration where your ellipse is centered and your line is horizontal.

If you want your ellipse in centered orientation, then you need to rotate the line about this center and use its parametric version first translated by {-xe,-ye} and then rotated by phi which is (refer to Link):

{u Cos[phi]+v Sin[phi]-xe,−u Sin[phi]+v Cos[phi]-ye}

Entering this into the original line equation y=k x + d, solving for y and using the rotation angle -ArcTan[k] yields a horizontal line

Solve[k (u Cos[t] + v Sin[t] - xe) + d - ye == -u Sin[t] + 
  v Cos[t], v][[1, 1]] /. {u -> x, v -> y, t -> -ArcTan[k]} // Simplify

with

yl=(d - k xe - ye)/Sqrt[1 + k^2]

as the y-coordinate independent of x as expected.

The rotation angle is simply the negative of your angle phi between your line and the x-axis which is

phi=ArcTan[x2 - x1, y2 - y1]=ArcTan[k]

thus yielding the new parametric form of the Ellipse as

{a Cos[t] Cos[Theta-phi] - b Sin[t] Sin[Theta-phi], 
 a Cos[t] Sin[Theta-phi] + b Sin[t] Cos[Theta-phi]}

The intersection points are easy. They can be calculated from the transformed parametric form of the Ellipse with parameters t1 and t2 and then the differences between their x-coordinates is the length of the line L.

Now we have everything to calculate the area of the shaded region as the integral of the radius vector of the ellipse r=Sqrt[x^2+y^2] along the parameter interval {t1,t2} minus the area of the triangle (0 t1 t2) which is yl L / 2. The interval {t1,t2} is determined by the condition that the y-coordinate of the parametric ellipse equation must be equal to yl

Solve[a Cos[t] Sin[Theta-phi] + b Sin[t] Cos[Theta-phi] == yl, t] /. C[1] -> 0 // Simplify

The integral can be carried out analytically (it is actually an Elliptic integral)

 Integrate[ Sqrt[(a Cos[t] Cos[Theta-phi] - b Sin[t] Sin[Theta-phi])^2 + 
 (a Cos[t] Sin[Theta-phi] + b Sin[t] Cos[Theta-phi])^2] // Evaluate, {t, t1, t2}, 
 Assumptions -> {{a, b, t1, t2, Theta, phi} \[Element] Reals, t2 > t1,a>0,b>0}]

yielding the area

a (-EllipticE[t1, 1 - b^2/a^2] + EllipticE[t2, 1 - b^2/a^2]) -  yl L /2

which is independent of the rotation angle Theta.

Answer 2

Rainer showed that we can use our calculus skills to set up the proper integral and evaluate it analytically.

But we can also solve this problem using the Region functionality in Mathematica to solve the problem. Let's take an arbitrary ellipse and a couple of points, as stated in the OP

ellipse = TransformedRegion[
   Ellipsoid[{14, 14}, {5, 2}],
   RotationTransform[\[Pi]/3]
   ];
{p1, p2} = {{-2, 17}, {-8, 16}};
Graphics[{Red, ellipse, Black, Line@{p1, p2}, Green, PointSize[Large],
   Point@{p1, p2}}, Frame -> True]

We can find the length of the line inside the ellipse by

ArcLength@RegionIntersection[InfiniteLine[{p1, p2}], ellipse]
N@%
(* (40 Sqrt[259 (-6425 + 3828 Sqrt[3])])/(2881 - 252 Sqrt[3]) *)
(* 3.77312 *)

To get the area, we form a HalfPlane from the line, visualized below

Graphics[{Red, ellipse, Black, Opacity[0.5], 
  HalfPlane[{p1, p2}, {0, -1}]}, Frame -> True]

N@
 Area@RegionIntersection[HalfPlane[{p1, p2}, {0, -1}], ellipse]
(* 3.81495 *)

It's easy to verify that the sum of the areas found by using the two half planes from this line add up to the total area of the ellipse.