# Cross section point of two conics curves

We have $A_i , B_i , C_i , D_i , E_i ,F_i, \ (i=1, 2)$.

We want to find $(u,v) \in \mathbb{R}^2$ satisfying

\begin{equation} A_1 u^2 + B_1 uv + C_1 v^2 + D_1 u + E_1 v +F_1 =0 \\ A_2 u^2 + B_2 uv + C_2 v^2 + D_2 u + E_2 v +F_2 =0. \end{equation}

Is there a general method to find exact solution (not an approximate solution)?

Take the resultant of the two equations with respect to one of the variables, say $v$. You'll get a quartic polynomial in $u$ whose coefficients are (complicated) polynomials in $A_1,\ldots,F_2$, but if $A_1,\ldots,F_2$ are real numbers, as you seem to suggest, then you'll get a quartic in $u$ with real coefficients. Then you can use the formula giving the exact solution to a quartic equation to find the four possible values of $u$. Similarly, you can find the four values of $v$. Then it's just a matter of matching which $u$ goes with which $v$. (You say your want to find the solutions in $\mathbb R$, but of course, sometimes there will only be solutions in $\mathbb C$.)