A = r(\cos \theta,\sin \theta); \quad r \to \infty

L_D(C) = D

E = sC + (1-s)D = [\underbrace{sI_2 + (1-s)L_D}_{L_E}](C)}

Known result:
The image of a conic curve under a non singular
linear transfomation is a conic of the same kind

v_1 \mbox{ eigenvector for eigenvalue }\mu_1 = -1

v_2\mbox{ eigenvector for eigenvalue \mu_2 = 1}

\mbox{Intersecting points: }span_{v_2} \cap c

a = 8 (slider from 0 to 10, step 0.5)
b = 4 (slider from 0 to 10, step 0.5)
c: x^(2) / a^(2) + y^(2) / b^(2) = 1
text1: "c: \frac{x^2}{" + (LaTeX[a]) + "^2} + \frac{y^2}{" + (LaTeX[b]) + "^2} = 1"
r = 20 (slider from 20 to 400, step 10)
theta = 0.39269908169872414 (slider from 0 to 6.283185307179586, step π / 16)
A: ((r * cos(theta)), (r * sin(theta)))
l = Segment((0, 0), A)
xhat(t) = (a * cos(t))
yhat(t) = (b * sin(t))
C = Point(c)
m = Line(A, C)
D = Intersect(c, m, 2)
s = 1.5 (slider from 0 to 5, step 0.25)
E: (s * C) + ((1 - s) * D)
k(t) = (yhat(t) - y(A)) / (xhat(t) - x(A))
a2(t) = ((a^(2) * k(t)^(2)) + b^(2)) / ((a^(2) * b^(2)))
a1(t) = (((2 * y(A)) * k(t)) - ((2 * x(A)) * k(t)^(2))) / b^(2)
x1(t) = ((-a1(t))) / a2(t) - xhat(t)
y1(t) = y(A) + (k(t) * (x1(t) - x(A)))
x_{E}(t) = (s * xhat(t)) + ((1 - s) * x1(t))
y_{E}(t) = (s * yhat(t)) + ((1 - s) * y1(t))
g = CurveCartesian(x_{E}(t), y_{E}(t), t, 0, (2 * π))
L_D: (1 / ((a^(2) * sin(theta)^(2)) + (b^(2) * cos(theta)^(2))) * {{a^(2) - ((b^(2) + a^(2)) * cos(theta)^(2)), (-((((2 * a^(2)) * cos(theta)) * sin(theta))))}, {(((-2 * b^(2)) * cos(theta)) * sin(theta)), ((b^(2) + a^(2)) * cos(theta)^(2)) - a^(2)}})
I_2: {{1, 0}, {0, 1}}
L_E: (s * I_2) + ((1 - s) * L_D)
c' = ApplyMatrix(L_E, c)
v_1: (3 * (1, ((-(cos(theta)^(2) - 1))) / ((cos(theta) * sin(theta)))))
v_{2}: (-3 * (1, ((-((b^(2) * cos(theta))))) / ((a^(2) * sin(theta)))))
span_{v1} = Line((0, 0), v_1)
span_{v2} = Line((0, 0), v_{2})
V_1, V_2, V_3, V_4 = Vertex(c')
axis_1 = Vector(V_2)
axis_2 = Vector(V_4)