Can all points in the plane be represented like this? 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! 
 A: 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 $-1$).
A: Let's play around and see what happens.
I write $a$ for $\alpha$,
$c$ for $\cos \alpha$,
and $s$ for $\sin \alpha$
cause I'm lazy.
We have
$x = t c + s$
and 
$y = t s + c$.
$x^2+y^2 = t^2 c^2+2tcs + s^2 + t^2 s^2 + 2tcs + c^2
= t^2+1+4tsc$.
$xy = t^2 cs + tc^2+t s^2 + sc
= cs(t^2+1)+t$.
We can solve for $cs$ in each of these.
$cs = (x^2+y^2-t^2-1)/(4t)$
and
$cs = (xy-t)/(t^2+1)$
so
$(x^2+y^2-t^2-1)/(4t) = (xy-t)/(t^2+1)$.
Now we can solve for $t$.
$4t(xy-t) = (t^2+1)(x^2+y^2-t^2-1)$
or
$4txy-4t^2 = (t^2+1)(x^2+y^2) - (t^2+1)^2
= t^2(x^2+y^2)+(x^2+y^2)-t^4-2t^2-1
$
so
$t^4-t^2(x^2+y^2+2)+4txy+1=0$.
This is a quartic, unfortunately,
and I don't see an obvious root.
We can also do this:
$xs-yc = (tcs + s^2)-(tsc+c^2)
= s^2-c^2
= 2s^2 - 1
$,
so
$yc = xs-2s^2+1$.
Squaring this,
and using $c^2 = 1-s^2$,
we get another quartic for $s$.
In other words,
we can get equations for
$t$
and $\sin \alpha$,
but they are quartics.
A: Here is an attempt at a visual interpretation.
The equation is equivalent to 
$$ \frac{ x - \sin \alpha} { \cos \alpha} = t = \frac{ y - \cos \alpha} { \sin \alpha},$$
$$ \frac{ y - \cos \alpha } { x - \sin \alpha} = \frac{ \sin \alpha} { \cos \alpha}$$
What this means, is that the slope of the point $(x,y)$ to the point $A=(\sin \alpha, \cos \alpha)$ on the unit circle, is $\tan \alpha$. 
