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Represent the quadratic form by the symmetric matrixThe spectral theorem for real symmetric matrices gives an orthogonal matrix $Q$ and real eigenvalues ${\lambda }_1,{\lambda }_2$ such that $Q^{T}MQ=\mathrm{diag}({\lambda }_1,{\lambda }_2)$. In the orthogonal change of coordinates $\mathbf{x}=Q\mathbf{u}$, the equation becomes ${\lambda }_1{u}_1^2+{\lambda }_2{u}_2^2=1$.

If both eigenvalues are positive, this is an ellipse with semiaxes ${\lambda }_1^{-1/2}$ and ${\lambda }_2^{-1/2}$. Conversely, a zero eigenvalue makes the set either empty or unbounded along that coordinate, while two negative eigenvalues give the empty set. Opposite signs give an unbounded hyperbola. A negative eigenvalue together with a zero one also gives the empty set. None of those cases is an ellipse. Therefore the exact condition is that $M$ be a symmetric positive-definite matrix.

The sum and product of its eigenvalues are its matrix trace and determinant:Two real numbers are both positive precisely when their sum and product are positive. ThusAn orthogonal transformation preserves area. Using i gives the quadratic-form ellipse area

Eliminate $z$ using the plane equation. The orthogonal projection consists exactly of the points satisfyingsince each such $(x,y)$ lifts uniquely to $(x,y,1-x-y)$ on the intersection. Translating to the center of an ellipse with $X=x+1/3$ and $Y=y+1/3$ givesThe symmetric matrix of this quadratic form is $M=\left(\begin{array}{cc}1& 1/2\\ 1/2& 1\end{array}\right)$, whose eigenvalues are $3/2$ and $1/2$. Thus it is a symmetric positive-definite matrix, and the projected curve is an ellipse. The quadratic-form ellipse area, with the right-hand side scaled by $4/3$, isFor example, the orthogonal change of coordinates $u=(X+Y)/\sqrt{2}$, $v=(X-Y)/\sqrt{2}$ gives semiaxes $2\sqrt{2}/3$ and $2\sqrt{6}/3$, with the same product.

Write $c=\cos \theta$, $s=\sin \theta$. Since $\theta$ is constant, the multivariable chain rule gives ${\partial }_x=c{\partial }_u-s{\partial }_v$ and ${\partial }_y=s{\partial }_u+c{\partial }_v$. Applying these operators twice to a twice continuously differentiable function givesThe mixed terms cancel and $c^2+s^2=1$, soThis is rotational invariance of the Laplacian: an orthogonal change of coordinates preserves the trace of the Hessian matrix. A nonconstant angle would introduce additional derivatives and is not covered by this constant rotation calculation.

By the spectral theorem for real symmetric matrices, choose an orthonormal eigenbasis ${\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3$ corresponding to $6,\sqrt{3},-\sqrt{3}$, with ${\mathbf{u}}_1=(1,1,1{)}^T/\sqrt{3}$. Write $P=[{\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3]$ and $\mathbf{x}=P\mathbf{\xi }$. This orthogonal change of coordinates givesThe plane constraint is therefore ${\xi }_1=0$. On that plane the quadratic form vanishes exactly whenConsequently the entire set isThe two direction vectors are nonzero and not parallel, since ${\mathbf{u}}_2,{\mathbf{u}}_3$ are linearly independent. An invertible orthogonal transformation maps each full line to a full line. Thus the set consists of precisely two infinite lines through the origin.

For a real symmetric matrix $A$, find its real eigenvalues from $\det (A-\lambda I)=0$, then solve $(A-\lambda I)v=0$ for the corresponding eigenspaces. Distinct eigenspaces are orthogonal. Within each repeated eigenspace, apply the Gram-Schmidt process and normalize to obtain an orthonormal basis of eigenvectors.

Put these vectors in the columns of $O$. The spectral theorem for real symmetric matrices guarantees a full basis, so $O^{T}O=I$ andThis is the orthogonal diagonalization of $A$ and the associated orthogonal change of coordinates.

The orthogonal change of coordinates preserves the sphere equation, giving $u^2+v^2+w^2=4$. Subtract it from the spheroid equation in part (b):Thus the intersection consists of the two circlesTheir common radius is $\boxed{\sqrt{5/8}=\sqrt{10}/4}$. Transforming their centers back to the original coordinates givesTheir planes have equations $x_1+x_2=\pm 3\sqrt{3}/2$, each with normal vector $(1,1,0)$. The difference ${\mathbf{c}}_{+}-{\mathbf{c}}_{-}$ is parallel to this normal, proving that the line joining the centers is perpendicular to both planes. This is the intersection of a prolate spheroid and a concentric sphere in the regime between the short and long semiaxes.

The orthogonal change of coordinates preserves distance to the origin. On the required quadric surface,Thus the squared distance is at least $1/4$. Equality holds precisely when $u=v=0$, and the constraint then gives $w=\pm 1/2$. Using $e_w=(1,0,-3)/\sqrt{10}$ from part (b), the nearest points in the original coordinates areSubstitution checks the constraint directly: $5x^2-3z^2+6xz=(5-27-18)/40=-1$. The inequality proves these are global nearest points, rather than merely stationary candidates.

For a two-dimensional symmetric positive-definite matrix $M$ and $R>0$, the region ${\mathbf{x}}^{T}M\mathbf{x}\le R$ is a filled ellipse of area $\pi R/\sqrt{\det M}$. By the spectral theorem for real symmetric matrices, an orthogonal change of coordinates gives positive eigenvalues ${\lambda }_1,{\lambda }_2$ and semiaxes $\sqrt{R/{\lambda }_1}$ and $\sqrt{R/{\lambda }_2}$. Their product and the area of an ellipse prove the formula. Translating the center of an ellipse leaves the area unchanged.