According to the math book, 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. A post suggested uses Singular value decomposition to find the coefficients (How to find an ellipse, given five points?) but I dont know how. Someone can explain to me the steps to obtain the coefficients.
Hint: In a webpage I read the last column vector of V (Right Singular Vectors) are the coefficient a,b,c,d,f,g but these values are not exactly what I was expecting.
Hint: If you put your five points ($(x_1,y_1),\dots,(x_5,y_5)$ ) in $ax^2+bxy+cy^2+dx+ey+f=0$
You will get 5 equations
$$ax_1^2+bx_1y_1+cy_1^2+dx_1+ey_1+f=0$$ $$ax_2^2+bx_2y_2+cy_2^2+dx_2+ey_2+f=0$$ $$ax_3^2+bx_3y_3+cy_3^2+dx_3+ey_3+f=0$$ $$ax_4^2+bx_4y_4+cy_4^2+dx_4+ey_4+f=0$$ $$ax_5^2+bx_5y_5+cy_5^2+dx_5+ey_5+f=0$$ You have six unknown $a,b,c,d,e,f$ but you can only have five equations with given five points. Other required sixth equation can be gotten by any point (x,y) on the ellipse .
Thus you can write six equations as: $$ax^2+bxy+cy^2+dx+ey+f=0$$ $$ax_1^2+bx_1y_1+cy_1^2+dx_1+ey_1+f=0$$ $$ax_2^2+bx_2y_2+cy_2^2+dx_2+ey_2+f=0$$ $$ax_3^2+bx_3y_3+cy_3^2+dx_3+ey_3+f=0$$ $$ax_4^2+bx_4y_4+cy_4^2+dx_4+ey_4+f=0$$ $$ax_5^2+bx_5y_5+cy_5^2+dx_5+ey_5+f=0$$
If you try to solve the six unknown equation , the equation system can only have a solution if
$$\left|\begin{array} &x^2 & xy & y^2 & x & y & 1 \\ x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1 \\ x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1 \\ x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1 \\ x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1 \\ x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1 \end{array}\right| = 0$$
because right side of the six equations are zero.
If determinant is not zero, there is no solution of the six equations. Thus the determinant must be zero .
SVD is overkill here.
Looking at the determinant form of the equation, the coefficient you are looking for are just the cofactors corresponding to the first row.
