Osculating circle of ellipse

Question

The osculating circle of ellipse at point $A$ interects the ellipse at $A,D$. I want to prove the tangent line $AB$ and the line $AD$ form equal angles with axis of the ellipse.

My attempt: The osculating circle of ellipse$$x^2/a^2+y^2/b^2=1\tag1$$at $A(a\cos(t),b\sin(t))$ is $$\left(x-\frac{\left(a^2-b^2\right) \cos ^3(t)}{a}\right)^2+\left(y+\frac{\left(a^2-b^2\right) \sin ^3(t)}{b}\right)^2=\frac{\left(a^2 \sin ^2(t)+b^2 \cos ^2(t)\right)^3}{a^2 b^2}\tag2$$ Solving the equations $(1),(2)$ to get the intersections $A$ and $D$. Then I find the coordinates of $D$ is $(a \cos (3 t),-b \sin (3 t))$.

Then I calculate the slope of $AD$ $$\frac{b \sin (t)--b \sin (3 t)}{a \cos (t)-a \cos (3 t)}=\frac{b \cot (t)}{a}$$ and the slope of $AB$ is $$\frac{\frac{d}{dt}b\sin(t)}{\frac{d}{dt}a\cos(t)}=-\frac{b \cot (t)}{a}$$ so $AB,AD$ form equal angle with $y$ axis (or $x$ axis).

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Solving equations $(1),(2)$ took very long computation, but in the end I find that the solution of $D$ is quite simple: $(a \cos (3 t),-b \sin (3 t))$. So I wonder if there is a better approach to this problem.

Answer 1

Theorem 1 in this article is

Let $A, B, C, D$ be four distinct points on $C$. Then $A, B, C, D$ are concyclic if and only if $[AC, BD] \cong\text{Sim}(C)$, that is, the axes of $C$ are parallel to the bisectors of the angles formed by the lines $AC$ and $BD$.

Let the points $B,C$ approach the point $A$, then the circle approaches the osculating circle in this problem, so $AC$ approaches the tangent line at $A$ and the line $BD$ approaches $AD$, so the limit of these two lines still form equal angle with y axis (or x axis). QED.