How to find an ellipse, given five points?

Is there a way to find the parameters $$A, B, \alpha, x_0, y_0$$ for the ellipse formula $$\frac{(x \cos\alpha+y\sin\alpha-x_0\cos\alpha-y_0\sin\alpha)^2}{A^2}+\frac{(-x \sin\alpha+y\cos\alpha+x_0\sin\alpha-y_0\cos\alpha)^2}{B^2}=1$$ given five points of the ellipse?

Here is one way to determine the equation of an ellipse given 5 points. (From there, you can work out the parameters that you want.)

Every ellipse has the form $ax^2+bxy+cy^2+dx+ey+f=0$. We'll find $a,b,c,d,e,f$ given 5 points.

Let $(p_1,q_1),\dots,(p_5,q_5)$ be your 5 points. Consider the following linear equations in the variables $a,b,c,d,e,f$: $ax^2+bxy+cy^2+dx+ey+f=0$, $ap_1^2+bp_1q_1+cq_1^2+dp_1+eq_1+f = 0$,...,$ap_5^2+bp_5q_5+cq_5^2+dp_5+eq_5+f = 0$. Note that we're treating $x^2, xy,y^2,x,y,1$ as coefficients in the first equation. Similarly, $p_i^2,p_iq_i, q_i^2, p_i,q_i,1$ are coefficients in the remaining 5 equations.

Since all 5 points lie on the same ellipse, the above linear equations must have common solution. From this, we see that the equation of the ellipse is given by setting the determinant of the matrix of coefficients of the above 6 equations to $0$.

What jrajchgot said, but written as a determinant... The conic section passing through the five points $(p_1,q_1),\dots,(p_5,q_5)$ has equation $ax^2+bxy+cy^2+dx+ey+f=0$ which may be written as the determinant equation $$\left|\begin{array} &x^2 & xy & y^2 & x & y & 1 \\ p_1^2 & p_1q_1 & q_1^2 & p_1 & q_1 & 1 \\ p_2^2 & p_2q_2 & q_2^2 & p_2 & q_2 & 1 \\ p_3^2 & p_3q_3 & q_3^2 & p_3 & q_3 & 1 \\ p_4^2 & p_4q_4 & q_4^2 & p_4 & q_4 & 1 \\ p_5^2 & p_5q_5 & q_5^2 & p_5 & q_5 & 1 \end{array}\right| = 0$$