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.