Rotating a natural axis plane z,w to a cartesian plane x,y for a rotated parabola

I am trying to get the equation of a parabola containing points ($0,0$), ($1,1$), tilted at a 45 degree angle. I know there are a multiple such parabolas, but I am only trying to get one.

To do this, I constructed a new set of axes, {$z,w$}, to represent the natural axes of the parabola. The $z$ axis represents the $x$ axis rotated 45 degrees counterclockwise, and the $w$ axis represents the $y$ axis rotated 45 degrees counterclockwise.

So on the {$z,w$} axis, the cartesian coordinate ($0,0$) is mapped to the z,w coordinate of ($0$,$0$), and the cartesian coordinate ($1,1$) is mapped to ($\sqrt{2}$,$0$). These two (z,w) mappings are the roots of the parabola in the {z,w} plane.

So the x-intercepts of the parabola in the {$z$,$w$} axis are $(\sqrt{2}$$,$$0$), and ($0,0$). Since there are an infinite such parabolas, let's define the vertex of the parabola to be ($\frac{\sqrt{2}}{2}$,$\frac{\sqrt{2}}{2}$). So our equation of our parabola is: $$(z-\frac{\sqrt{2}}{2})^2=-4a(w-\frac{\sqrt{2}}{2})$$($-4a$ because the parabola is facing down)

To find $a,$ we can plug in a point on the parabola, say ($\sqrt{2},0$) to get $a=\frac{\sqrt{2}}{8}$.

So our equation of the parabola in the $z$ and $w$ axes plane is: $(z-\frac{\sqrt{2}}{2})^2=-\frac{\sqrt{2}}{2}*(w-\frac{\sqrt{2}}{2})$

Relating $z$ with $z$ and $w$ with $y$, we have: $\frac{x}{z}=$sin$45^o$ $\rightarrow z = x\sqrt{2}$ and $\frac{y}{w}=$cos$45^o \rightarrow w = y\sqrt{2}$

So now we replace $w$ in terms of $y$ and $z$ in terms of $x$: $$(x\sqrt{2}-\frac{\sqrt{2}}{2})^2 = -\frac{\sqrt{2}}{2} * (y\sqrt{2}-\frac{\sqrt{2}}{2})$$ However, when I try to plug in (x,y) = (1,1), it does not work. I am a negative sign off. What am I doing wrong? I think I made a computation error with relating the $(z,w)$ to $(x,y)$ in the second to last step, because the parabolic equation in terms of $w$ and $z$ works with the three ($w,z$) points.

You've got the wrong formula for mapping $(z,w)$ coordinates into $(x,y)$ coordinates.
By your construction, you rotate a point $(x,y)$ in $x,y$-space by $\frac\pi4$ clockwise about the origin to get its coordinates in $z,w$-space. So if a point $(x,y)$ in $x,y$-space is expressed in polar coordinates as $(r\cos\theta, r\sin\theta)$, then the coordinates in $z,w$-space of that point will be $(z,w)=(r\cos(\theta-\frac\pi4), r\sin(\theta-\frac\pi4))$. Using the addition formulas for sine and cosine, this simplifies to $$z=\frac{y+x}{\sqrt 2},\qquad w=\frac{y-x}{\sqrt 2}.\tag1$$ Plugging (1) into the $z,w$-space equation of your parabola and simplifying, we get the $x,y$-space equation $$y-x-1=-(x+y-1)^2.\tag2$$ The points $(0,0)$ and $(1,1)$ satisfy (2). Here's a plot of the equation.