Suppose we have a 2$$-d$$ quadratic form $$Q : \mathbb{R}^2 \to \mathbb{R}$$. How can we compute its determinant?

This problem is a little under-specified. In fact, we can always pick a basis for $$\RR^2$$ so that our quadratic form is diagonal with diagonal entries $$0$$ or $$\pm 1$$. So there is always some basis where our quadratic form has determinant $$0$$ or $$\pm 1$$.

But usually, we have a choice of inner product on $$\mathbb{R}^2$$, and if we restrict to orthonormal bases with respect to this inner product then $$Q$$ does indeed have well-defined eigenvectors, and a well-defined trace and determinant. And once we have the eigenvectors, the trace and determinant are easy to evaluate. If $$e_1, e_2$$ are the eigenvectors with eigenvalues $$\lambda_1, \lambda_2$$, then the trace of $$Q$$ is given by $\tr Q = \lambda_1 + \lambda_2 = Q(e_1) + Q(e_2),$ and the determinant of $$Q$$ is given by $\det Q = \lambda_1 \lambda_2 = Q(e_1)Q(e_2).$

All this eigenvector business is actually not needed for computing the trace: we can pick any orthonormal vectors $$X, Y \in \mathbb{R}^2$$, and the trace is given by $$Q(X) + Q(Y)$$.

However, the determinant is trickier. In general, $$\det Q$$ is not equal to $$Q(X)Q(Y)$$ for a general pair of orthonormal vectors $$X$$ and $$Y$$. To see what goes wrong, we can write out $$Q$$ as a matrix in the $$X, Y$$ basis. The diagonal entries of this matrix are simply $$Q(X)$$ and $$Q(Y)$$. To obtain the off-diagonal entries, we can apply the polariziation identity (essentially a fancy name for the fact that $$xy = \tfrac{1}{2}(x+y)^2 - \tfrac{1}{2}x^2 - \tfrac{1}{2}y^2$$). Now that we have the matrix entries for $$Q$$, we can apply the usual formula that $$\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = a d - b c$$ and we find that $\det Q = Q(X) Q(Y) - \tfrac{1}{4}\left(Q(X+Y) - Q(X) - Q(Y)\right)^2.$

This formula simplifies a little if we define a new unit vector $$Z := \tfrac{1}{\sqrt{2}}\left(X + Y\right)$$ which lies half way in between $$X$$ and $$Y$$. Then we can do a little bit of algebra to find that \begin{aligned} \det Q &= Q(X) Q(Y) - \left(Q(Z) - \tfrac{1}{2}\left(Q(X) + Q(Y)\right)\right)^2\\ &= Q(X)Q(Y) - Q(Z)^2 - \tfrac{1}{4}\left(Q(X)+Q(Y)\right)^2 + Q(Z)\left(Q(X) + Q(Y)\right)\\ &= \tfrac{1}{4}\left(Q(X) - Q(Y)\right)^2 + Q(Z) \left(Q(X) + Q(Y) - Q(Z)\right). \end{aligned}

We can simplify this formula even more by introducing the complementary unit vector $$W := \tfrac{1}{\sqrt{2}}\left(X - Y\right)$$. Note that $$Z, W$$ also form an orthonormal basis for $$\RR^2$$. Since we can use the any orthonormal basis of $$\RR^2$$ to compute the trace of $$Q$$, we know that $$Q(X) + Q(Y) = \tr Q = Q(Z) + Q(W)$$, and therefore the expression $$Q(X) + Q(Y) - Q(Z)$$ in our formula is simply equal to $$Q(W)$$. So in the end, the determinant of $$Q$$ is given by $\det Q = \tfrac{1}{4}\left(Q(X) - Q(Y)\right)^2 + Q(Z) Q(W),$ where $$X$$ and $$Y$$ form one orthonormal basis, and $$Z$$ and $$W$$ form another orthonormal basis rotated by $$45^\circ$$.

As one last sanity check, note that if we take $$Z$$ and $$W$$ to be the eigenvectors $$e_1$$ and $$e_2$$, then $$Q(X) = Q(Y)$$, and we so find again that the determinant is given by $$\det Q = Q(e_1) Q(e_2)$$. But now we see that this is really a special case, and in general we need to add a correction term involving a pair of vectors rotated by $$45^\circ$$ from our starting vectors.