Quadric and tangents planes Let $Q$ be the quadratic $x^2 + 4xy - 2y^2 + 6z^2 + 2y +2z = 0$


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*Prove that $Q$ is a cone and find its vertex.

*Write the tangent plane $A$ to the cone in $(0,0,0)$ and say which kind of conic is the intersection of $Q$ with $A$.

*Find an ellipse and a parabola on the cone.


Am a bit ok for 1 and 2 but require some help for 3. For 1 we need the matrix representation of the cone  and the 2 differentiate with respect to $x$ and then plug in $(0,0,0)$.
Anybody with any different ideas... 

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*With $A=\begin{pmatrix}1&2&0\\2&-2&0\\0&0&6\end{pmatrix}$ and $K=\begin{pmatrix}0&2&2\end{pmatrix}$, the equation is 
${\bf x}^TA{\bf x}+K{\bf x}=0$. $A$ has eigenvalues $2,-3,6$ and the eigenvector matrix is $P=\begin{pmatrix}\frac{2\sqrt{5}}{5}&-\frac{\sqrt{5}}{5}&0\\\frac{\sqrt{5}}{5}&\frac{2\sqrt{5}}{5}&0\\0&0&1\end{pmatrix}$. 
${\bf x}=P{\bf x}'$ gives ${\bf x}'^TP^TAP{\bf x}'+KP{\bf x}'=0$ where we complete the squares and that $x''=x'+\frac{\sqrt{5}}{10}, y''=y'-\frac{2\sqrt{5}}{15}, z''=z'+\frac{1}{6}$ transforms the equation into $2x''^2-3y''^2+6z''^2=0$. The vertex should be $P\begin{pmatrix}-\frac{\sqrt{5}}{10}\\\frac{2\sqrt{5}}{15}\\ -\frac{1}{6}\end{pmatrix}=\begin{pmatrix}-\frac{1}{3}\\\frac{1}{6}\\ -\frac{1}{6}\end{pmatrix}$.

*To get the tangent plane at $(0,0,0)$ we find the gradient which is $\begin{pmatrix}0&2&2\end{pmatrix}$ and is a normal vector. $2y+2z=0$. Intersecting is considering the two equations a system, so we put $z=-y$ into the first equation and get: $(x+2y)^2$ a double line.

*To get an ellipse we need a plane which only intersects one nappe of the cone. $y=0$ should do. Indeed putting it into the equation we get $6x^2+36(z+\frac{1}{6})^2=1$.
To get a parabola we need a plane parallel to a plane which is tangential to the cone's surface. We found one in 2. So 2y+2z=2 should do.
