= Solution
{c}
The <quadratic form> is
$$
5x^2+6xy+5y^2=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}5&3\\3&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}.
$$
The <symmetric matrix> has <eigenvalues> $2,8$, with respective <orthogonal> <eigenvectors> $(1,-1)^T,(1,1)^T$. Normalize them to obtain the <orthonormal basis>
$$
\boxed{\mathbf e_1=\frac{\mathbf i-\mathbf j}{\sqrt2},\qquad \mathbf e_2=\frac{\mathbf i+\mathbf j}{\sqrt2}.}
$$
The given <change of basis> means
$$
x=\frac{X+Y}{\sqrt2},\qquad y=\frac{-X+Y}{\sqrt2}.
$$
Substitution, or <orthogonal diagonalization of a real symmetric matrix>, gives
$$
2X^2+8Y^2=16,\qquad \boxed{\frac{X^2}{8}+\frac{Y^2}{2}=1.}
$$
Thus the <semi-major and semi-minor axes> have lengths $2\sqrt2$ and $\sqrt2$. The <eccentricity> and enclosed <area> are
$$
\boxed{e=\frac{\sqrt3}{2},\qquad A=4\pi.}
$$
The <orthogonal> <change of basis> preserves <area>, so this is also the <area> in the original <Cartesian coordinates>.
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