To which of the seventeen standard quadrics do these two equations reduce?

Question

To which of the seventeen standard quadrics (https://mathworld.wolfram.com/QuadraticSurface.html) do these two equations reduce? \begin{equation} Q_1^2+3 Q_2 Q_1+\left(3 Q_2+Q_3\right){}^2 = 3 Q_2+2 Q_1 Q_3. \end{equation} \begin{equation} -9 Q_2-6 Q_3+3 \left(Q_1^2+\left(3 Q_2+4 Q_3-1\right) Q_1+9 Q_2^2+4 Q_3^2+6 Q_2 Q_3\right) = 0. \end{equation} Further, what are the associated transformations needed to accomplish the reductions?

This is a "distilled" form of a previous more expansive question https://mathoverflow.net/questions/359459/interpret-certain-expressions-in-terms-of-classical-quadratic-surfaces

Answer 1

This is the simple way, uses rational numbers only. The alternative is to translate coordinates around the single critical point(if there is one and only one point where the gradient is the zero vector). Then, find the eigenvalues and construct an orthogonal matrix $P,$ leading to $P^T H P = D.$

your first one comes out, in letters xyz1, as $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 3 }{ 2 } & 1 & 0 & 0 \\ - 1 & \frac{ 2 }{ 3 } & 1 & 0 \\ 0 & - \frac{ 2 }{ 9 } & - \frac{ 1 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & \frac{ 27 }{ 2 } & 0 & 0 \\ 0 & 0 & - 6 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 2 } & - 1 & 0 \\ 0 & 1 & \frac{ 2 }{ 3 } & - \frac{ 2 }{ 9 } \\ 0 & 0 & 1 & - \frac{ 1 }{ 3 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & 3 & - 2 & 0 \\ 3 & 18 & 6 & - 3 \\ - 2 & 6 & 2 & 0 \\ 0 & - 3 & 0 & 0 \\ \end{array} \right) $$

which means $$ \small \left(x + \frac{3}{2}y -z \right)^2 + \frac{27}{4} \left(y + \frac{2}{3}z - \frac{2}{9}\right)^2 - 3 \left(z - \frac{1}{3}\right)^2 = x^2 + 9 y^2 + z^2 + 6yz - 2zx+3xy -3y $$

This is an actual cone, cross section elliptic, vertex at $x = 1/3 \; , \; \; y=0 \; , \; \; z=1/3 \; . \;$

Answer 2

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 3 }{ 2 } & 1 & 0 & 0 \\ 2 & 0 & 0 & 1 \\ - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 9 } & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 6 & 0 & 0 & 0 \\ 0 & \frac{ 81 }{ 2 } & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 3 }{ 2 } & 2 & - \frac{ 1 }{ 2 } \\ 0 & 1 & 0 & - \frac{ 1 }{ 9 } \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrrr} 6 & 9 & 12 & - 3 \\ 9 & 54 & 18 & - 9 \\ 12 & 18 & 24 & - 6 \\ - 3 & - 9 & - 6 & 0 \\ \end{array} \right) $$

This one has a nonzero constant involved, so no node.

$$ 3\left(x + \frac{3}{2}y +2z - \frac{1}{2}\right)^2 + \frac{81}{4} \left(y - \frac{1}{9}\right)^2 - 1 = 3x^2 + 27 y^2 + 12z^2 + 18yz +12zx+9xy-3x -9y -6z $$

This is an elliptic CYLINDER. Indeed, if we call the original polynomial $f(x,y,z),$ degree 2 but not homogeneous, we find $$ f( x_0 + 2t, y_0, z_0 - t) $$ is independent of $t$

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