= Solution
{c}
Represent the <quadratic form> by the <symmetric matrix>
$$
M=\begin{pmatrix}\alpha&\beta/2\\\beta/2&\gamma\end{pmatrix},\qquad
\mathbf x^TM\mathbf x=1.
$$
The <spectral theorem for real symmetric matrices> gives an <orthogonal matrix> $Q$ and real <eigenvalues> $\lambda_1,\lambda_2$ such that $Q^TMQ=\operatorname{diag}(\lambda_1,\lambda_2)$. In the <orthogonal change of coordinates> $\mathbf x=Q\mathbf u$, the <equation> becomes $\lambda_1u_1^2+\lambda_2u_2^2=1$.
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>:
$$
\lambda_1+\lambda_2=\alpha+\gamma,\qquad
\lambda_1\lambda_2=\alpha\gamma-\frac{\beta^2}{4}.
$$
Two real numbers are both positive precisely when their sum and product are positive. Thus
$$
\boxed{\alpha+\gamma>0,\qquad 4\alpha\gamma>\beta^2.}
$$
An <orthogonal transformation> preserves <area>. Using <2/a/i> gives the <quadratic-form ellipse area>
$$
\boxed{\mathcal A=\frac{\pi}{\sqrt{\det M}}=\frac{2\pi}{\sqrt{4\alpha\gamma-\beta^2}}.}
$$
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