Reduce to canonical form and identify…
Reduce to canonical form and identify: $5x^2+6y^2+7z^2-4xy+4yz-10x+8y+14z-6=0 $.
Explanation of Method
Reducing a central quadric to canonical form: eigenvalues, center, constant.
Step 1: Eigenvalues
\[ A=\begin{pmatrix}5&-2&0\\-2&6&2\\0&2&7\end{pmatrix},\qquad \lambda^3-18\lambda^2+99\lambda-162=(\lambda-3)(\lambda-6)(\lambda-9). \]
The corresponding eigenvectors are $\{2,2,-1\}$, $\{2,-1,2\}$ and $\{1,-2,-2\}$, which are mutually orthogonal.
Step 2: Center and constant
Solve $A\mathbf x_0=-\tfrac12\mathbf b=(5,-4,-7)^T$. This gives $\mathbf x_0=(1,0,-1)$. The new constant is
\[ d'=d+\tfrac12\mathbf b^T\mathbf x_0=-6+\tfrac12(-10-14)=-18 . \]
Step 3: Canonical form
\[ 3\xi^2+6\eta^2+9\zeta^2=18 . \]
\[\frac{\xi^2}6+\frac{\eta^2}3+\frac{\zeta^2}2=1\]
This is an ellipsoid with center $(1,0,-1)$ and semi-axes $\sqrt6$, $\sqrt3$, $\sqrt2$ along the eigenvectors above. A further scaling of the axes (an affine map) turns it into the unit sphere $\xi'^2+\eta'^2+\zeta'^2=1$.
- $\sqrt6\ \ \{2,2,-1\}$
- $\sqrt3\ \ \{2,-1,2\}$
- $\sqrt2\ \ \{1,-2,-2\}$
Result
\[\boxed{\,\frac{\xi^2}6+\frac{\eta^2}3+\frac{\zeta^2}2=1\,}\]