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.
The flattening factor is given by $\;f = 1 - \cfrac ba$.
A closely related term you might be interested in is the eccentricity of an ellipse, usually denoted $e$ or $\varepsilon$. Eccentricity in general represents ratio of the distance between the two foci, $2h$, to the length of the major axis, $2a$: $$e = \dfrac{2h}{2a} = \dfrac ha$$
where the distance between a focus and the center is given by $\;h = \sqrt{a^2-b^2}.$
In fact, we can represent eccentricity in terms of $a, b$:
$$e=\varepsilon=\sqrt{\frac{a^2-b^2}{a^2}} =\sqrt{1-\left(\frac{b}{a}\right)^2} = \frac ha$$
For an ellipse, the eccentricity is $0 < e < 1$. Eccentricity is zero when the foci coincide with the center point, i.e, the figure is a circle. As eccentricity increases, the shape gets more elongated (stretched/flattened): the closer to $1$ it gets, the flatter the ellipse.
Best Answer
Hint Substitution shows that the reflections $(x, y, z) \mapsto (-x, y, z)$ and $(x, y, z) \mapsto (x, -y, z)$ fix the plane and cylinder and thus the ellipse and thus the major and minor axes.