Can all points in the plane be represented like this?

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 $-1$).

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.

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 $\tan \alpha$.