How can I classify a conic given by $x^2+xy+3y^2-4x=0$ and $x+y+z-2=0$? I need help to know how can I classify this type of conic that has two equations.
Consider a conic : 
 $$
\left\{ 
\begin{array}{r}
x^2+xy+3y^2-4x=0 \\ 
x+y+z-2=0 \\ 
\end{array}
\right. 
$$
I know how to classify a conic with one equation by applying matrices and translation but how can i deal with this one ? 
 A: Each equation is the equation of a surface in $xyz$ space: the first is an elliptic cyclinder, the second a plane.  The intersection of the two is an ellipse.

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A: If you work in homogeneous coordinates, you can represent a quadric surface with a symmetric matrix $Q$: the equation of the quadric is them $(x,y,z,1)\,Q\,(x,y,z,1)^T=0$. In a similar way, if you parameterize a plane as $\mathbf P+\lambda\mathbf U+\mu\mathbf V$, you can write this parameterization in matrix form: $$\begin{bmatrix}x\\y\\z\\1\end{bmatrix} = \begin{bmatrix}\mathbf U&\mathbf V&\mathbf P\\0&0&1\end{bmatrix}\begin{bmatrix}\lambda\\\mu\\1\end{bmatrix} = M\mathbf x.$$ Essentially, the choice of the two direction vectors and point $\mathbf P$ imposes a coordinate system on the plane. Substituting this into the quadric equation gives $$(M\mathbf x)^TQ(M\mathbf x)=\mathbf x^T(M^TQM)\mathbf x = 0,$$ which is the equation of a conic in the $(\lambda,\mu)$ coordinate system of the plane. You can then classify this conic by examining the matrix $C=M^TQM$. Note that we don’t need $\mathbf u$ and $\mathbf v$ to form an orthonormal basis for this coordinate system: they are related to one via an affine transformation, which doesn’t change the type of a conic.  
For this problem, we have $$Q=\begin{bmatrix}1&\frac12&0&-2\\\frac12&3&0&0\\0&0&0&0\\-2&0&0&0\end{bmatrix}$$ and can take $$M=\begin{bmatrix}1&1&2\\-1&0&0\\0&-1&0\\0&0&1\end{bmatrix}.$$ We then have $$C=M^TQM=\begin{bmatrix}3&\frac12&-1\\\frac12&1&0\\-1&0&-4\end{bmatrix}.$$ $\det C=-12$ but $C$ is traceless, so it must have two positive eigenvalues and one negative eigenvalue, making the curve an ellipse.
