Magnetic field in the left focus of an ellipse current loop

given a 2D ellipse in 3D space parametrized as $$x(t)=a\cos{t}$$, $$y(t)=b\sin{t}$$, I would like to find the magnetic field at point $$(-c,0,0)$$, $$c=\sqrt{a^2-b^2}$$. (The current $$I$$ is running counter-clockwise)

Using the Biot-Savart law: $$dB=\frac{\mu_0I}{4\pi}\frac{|d\vec{l}\times \vec{r}|}{r^3}$$ I make use of $$|d\vec{l}\times \vec{r}|=dl \cdot r \sin{\alpha}$$ and then I further relate $$dl$$ with $$dt$$ as $$dl=\sqrt{x'^2+y'^2}\ dt$$ (calculating the length of the ellipse). Now I get something like $$dB=\frac{\mu_0I}{4\pi}\int_{-\pi}^{\pi}\frac{\sin{\alpha}\sqrt{a^2-c^2\cos{t}}}{(c\cos{t}+a)^2}dt$$ which may or may not be correct, but mostly I'd like to know how to relate the sine of the angle $$\alpha$$ between the two vectors with the parameter $$t$$ so I can complete the integral. I can't quite see it.

\begin{align} \mathbf B (-c,0,0) &= \frac{\mu_0 I}{4\pi} \oint_C \frac {d\boldsymbol s \times (-c\, \mathbf i-\mathbf r)} {|-c\, \mathbf i-\mathbf r|^3} \\[5pt] &= \frac{\mu_0 I}{4\pi} \oint_C \frac {\mathbf r'(t) \times (-c\, \mathbf i-\mathbf r)} {|-c\, \mathbf i-\mathbf r|^3} \, dt \\[5pt] &= \frac{\mu_0 I}{4\pi} \, \mathbf k \int_0^{2\pi} \frac{b(a+c\cos t) \, dt}{[(c+a\cos t)^2+b^2\sin^2 t]^{3/2}} \\[5pt] &= \frac{\mu_0 I}{4\pi} \, \mathbf k \int_0^{2\pi} \frac{b(a+c\cos t)}{[(a+c\cos t)^2]^{3/2}} \, dt \\[5pt] &= \frac{\mu_0 I}{4\pi} \, \mathbf k \int_0^{2\pi} \frac{b}{(a+c\cos t)^2}\, dt \\[5pt] &= \frac{\mu_0 I}{2\pi} \, \mathbf k \left[ \frac{2a}{b^2} \tan^{-1} \left( \sqrt{\frac{a-c}{a+c}} \tan \frac{t}{2} \right)- \frac{c\sin t}{b(a+c\cos t)} \right]_0^\pi \\[5pt] &= \frac{\mu_0 aI}{2b^2} \,\mathbf k \end{align}
• Note that $$\dfrac{2b^2}{a}$$ is the latus rectum of the ellipse.