Equation of a parabola, given the vertex and the axis I get totally stuck in this exercise:
Consider the set of parabolas with vertex in $V=(1,1)$ and, as axis, the line $y=2x-1$. Write the equation of the generic parabola of that set.
Thank you for your help!

Consider a parabola $y-1=A(x-1)^2$ rotated clockwise by $\alpha$ about the point $(1,1)$.  
The equation of the rotated parabola is
$$v=Au^2$$
where 
$$\begin{align}
\left(\begin{array} .
u\\v\end{array}\right)
&=
\left(\begin{array}.\;\ \cos(-\alpha)&\sin(-\alpha)\\-\sin(-\alpha)&\cos(-\alpha)\end{array}\right)
\left(x-1\atop y-1\right)\\
&=\left(\begin{array}. \cos\alpha&-\sin\alpha\\\sin\alpha&\;\;\ \cos\alpha\end{array}\right)
\left(x-1\atop y-1\right)\\
&=\left(\begin{array}. \sin\theta&-\cos\theta\\\cos\theta&\;\;\ \sin\theta\end{array}\right)
\left(x-1\atop y-1\right)
&&\scriptsize(\alpha+\theta=\tfrac\pi 2)\\ 
\end{align}$$
and $\tan\theta=2$ (slope of $y=2x-1$).
Hence equation of rotated parabola is 
$$\begin{align}
(x-1)\cos\theta+(y-1)\sin\theta
&=A\big[(x-1)\sin\theta-(y-1)\cos\theta\big]^2\\
\tfrac 1{\sqrt{5}}(x-1)+\tfrac 2{\sqrt{5}}(y-1)
&=A\big[\tfrac 2{\sqrt{5}}(x-1)-\tfrac 1{\sqrt{5}}(y-1)\big]^2\\
\tfrac 1{\sqrt{5}}\big[(x-1)+2(y-1)\big]&=\tfrac 15 A\big[2(x-1)-(y-1)\big]^2\\
x+2y-3
&=\tfrac 1{\sqrt{5}}A\big[2x-y-1\big]^2\\
\color{red}{x+2y-3}
&\color{red}{=A'\big[2x-y-1\big]^2}\\
\end{align}$$
where $A, A'$ are constants.
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