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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.