Prove that there exist at most two parabolas passing through four given points.
I recently get this:
If I have five points $(x_i,y_i),i=1,2,\ldots, 5$, I can represent the conic section by this determinant. $$ \begin{vmatrix} {1}&x&y&{xy}&{x^2}&{y^2} \\ {1}&{x_1}&{y_1}&{x_1y_1}&{x_1^2}&{y_1^2} \\ {1}&{x_2}&{y_2}&{x_2y_2}&{x_2^2}&{y_2^2} \\ {1}&{x_3}&{y_3}&{x_3y_3}&{x_3^2}&{y_3^2} \\ {1}&{x_4}&{y_4}&{x_4y_4}&{x_4^2}&{y_4^2} \\ {1}&{x_5}&{y_5}&{x_5y_5}&{x_5^2}&{y_5^2} \end{vmatrix}=0 $$
So I think: if I have four points, in theory, I will get at most 2 parabolas. But what I only know is: for $$Ax^2+2Bxy+Cy^2+\cdots=0$$ $\Delta=B^2-AC=0$ means a parabola.
Can I prove what I thought with this determinant?
Here,I have a not so rigorous prove. Consider this two curves that pass through four given points.
$$ T_1:A_1x^2+B_1 xy+C_1y^2+D_1x+E_1y+F_1=0\\ T_2:A_2x^2+B_2 xy+C_2y^2+D_2x+E_2y+F_2=0 $$
then we write it like this.
$$ (λ_1A_1+λ_2A_2)x^2+(λ_1B_1+λ_2B_2)xy+(λ_1C_1+λ_2C_2)y^2+⋯=0 $$
(since $Dx,Ey,F$ is unimportant to think about.)
apply $B^2-4AC=0$
$$(λ_1B_1+λ_2B_2)^2-4(λ_1A_1+λ_2A_2)(λ_1C_1+λ_2C_2)=0$$
expand it ,and we will get
$$\tag{1}(XX)λ_1^2+(YY)λ_1λ_2+(ZZ)λ_2^2=0$$
it's a Homogeneous equation.
At most it has two solutions. That is to say,we get 2 pairs of $\lambda_1,\lambda_2$,so we get two parabolas.