TITLE: Can all points in the plane be represented like this?
QUESTION [12 upvotes]: Solving a task regaring affine geometry, I've come across a problem:
Is it  true that, for every point $(x,y)\in \mathbb{R}^2$, there  exist $t\in \mathbb{R}, \alpha\in[0,2\pi]$, such that 
$$x = t\cos\alpha + \sin \alpha \\ y = t\sin\alpha + \cos \alpha?$$
I don't have any idea how to prove this or find a counterexample. All hints appreciated!

REPLY [4 votes]: This can be written as
\begin{align}
&\left(\begin{array}{cr}
\tilde{x} \\ \tilde{y}
\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{cr}
1 &-1 \\ 1 & 1
\end{array}\right)\left(\begin{array}{cr}
x \\ y
\end{array}\right)=
\left(\begin{array}{cr}
t-1 & 0 \\ 0 & t+1
\end{array}\right)\left(\begin{array}{c}
\cos\left(\alpha+{\pi}/{4}\right) \\ \sin\left(\alpha+{\pi}/{4}\right)
\end{array}\right).\tag{1}
\end{align}
Since $(\tilde{x},\tilde{y})$ are obtained by rotation of $(x,y)$ by $\pi/4$, it suffices to show that any point in $\mathbb{R}^2$ can be represented by the right side of (1). This is equivalent to showing that for any point $P=(x,y)$ with $x,y\neq 0$ one can always find $t\neq \pm1$ such that the ellipse
$$\left(\frac{x}{t-1}\right)^2+\left(\frac{y}{t+1}\right)^2=1 \tag{2}$$
passes through $P$ (which is true - think on what happens to this ellipse when $t$ continuously decreases from $\infty$ to $1$, and then from $1$ to $-1$).