More explicitly, we have the decomposition
$$\begin{pmatrix}\cos\,t\\\cos\,t+\sin\,t\end{pmatrix}=\begin{pmatrix}\cos\,\lambda&-\sin\,\lambda\\\sin\,\lambda&\cos\,\lambda\end{pmatrix}\cdot\begin{pmatrix}\sqrt{1+\phi}\cos(t+\eta)\\\sqrt{2-\phi}\sin(t+\eta)\end{pmatrix}$$
where $\tan\,\lambda=\phi$, $\tan\,\eta=1-\phi$, and $\phi=\dfrac{1+\sqrt{5}}{2}$ is the golden ratio. You can check that both your original parametric equations and the new decomposition both satisfy the Cartesian equation $2x^2-2xy+y^2=1$. What the decomposition says is that your curve is an ellipse with axes $\sqrt{1+\phi}$ and $\sqrt{2-\phi}$, with the major axis inclined at an angle $\lambda$.
If we take the linear algebraic viewpoint, as suggested by Robert in the comments, what the decomposition given above amounts to is the singular value decomposition (SVD) of the shearing matrix; i.e.,
$$\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}\cos\,\lambda&-\sin\,\lambda\\\sin\,\lambda&\cos\,\lambda\end{pmatrix}\cdot\begin{pmatrix}\sqrt{1+\phi}&\\&\sqrt{2-\phi}\end{pmatrix}\cdot\begin{pmatrix}\cos\,\eta&\sin\,\eta\\-\sin\,\eta&\cos\,\eta\end{pmatrix}^\top$$
The SVD is in fact an excellent way to look at how a matrix transformation geometrically affects points: the two orthogonal matrices on the left and right can be thought of as rotation matrices, reflection matrices, or products thereof, and the diagonal matrix containing the singular values amounts to nothing more than a scaling about the axes of your coordinate system.
Perhaps added since this question was answered, Wikipedia has good information on this problem. There is an interesting geometric construction which contrasts with the algebraic solutions offered here: Rytz's construction.
(I have been told to add information to the answer rather than just posting links. Unfortunately as my rep is less than 50 I can't make comments yet)
The setting in which I found Rytz's construction useful was in drawing the elevation of a circle in a plan oblique projection. In this case, as in the other conjugate tangent problems that arise in parallel projection, the ellipse is tangent to the midpoints of the edges of a parallelogram. This is a slightly more constrained and regular situation than the diagram referenced in the original question, though a tangent parallelogram could easily be constructed around the ellipse shown in that image.
Rytz's construction is apparently the last refinement of a long series of solutions to this problem, starting with Pappus. It relies on the fact that conjugate diameters are affine images of perpendicular diameters of a circle. In particular, the perpendiculars from the foci to any tangent intersect the tangent on the auxiliary circle, the circle centered at the centre of the ellipse with the major axis as diameter. As I understand it Rytz's construction is a carefully minimized (in terms of number of steps) derivative of the earlier techniques, intended for practical use in drafting, etc.
Best Answer
For example, let's take the 3-dimensional vectors $$ V_1 = \pmatrix{1\cr 2\cr 1\cr},\ V_2 = \pmatrix{3\cr -3\cr 2\cr} $$ With these as columns we form the matrix $$A = \pmatrix{1 & 3\cr 2 & -3\cr 1 & 2\cr}$$ Now $$ A A^T = \pmatrix{1 & 3\cr 2 & -3\cr 1 & 2\cr} \pmatrix{1 & 2 & 1\cr 3 & -3 & 2\cr} = \pmatrix{10 & -7 & 9\cr -7 & 13 & 0\cr 9 & 0 & 13\cr}$$ which has eigenvalues $23, 13, 0$ corresponding to normalized eigenvectors $$ W_1 = \frac{1}{\sqrt{230}} \pmatrix{10\cr -7 \cr 9\cr},\ W_2 = \frac{1}{\sqrt{130}} \pmatrix{0 \cr 9\cr 7\cr},\ W_3 = \frac{1}{\sqrt{299}} \pmatrix{13\cr 7\cr -9\cr} $$ Note that these are orthogonal since $A A^T$ is symmetric. The major and minor axes have lengths $\sqrt{23}$ and $\sqrt{13}$ (the singular values, which are the square roots of the nonzero eigenvalues of $A A^T$) and directions $W_1$ and $W_2$ respectively.