Normalisation of $k[x,y]/(y^2-x^2(x-1))$ 
I am trying to figure out the normalisation of $k[x,y]/(y^2-x^2(x-1))$, for an algebraically closed field $k$. 

I can show that it is not normal and I have the information that the normalisation is $k[t]$, where $t=\frac{y}{x}$. I can't figure out how to prove that this is the normalisation i.e. that the fraction fields of $k[x,y]/(y^2-x^2(x-1))$ and $k[t]$ are the same, that $k[t]$ is integrally closed and that $k[t]$ is an integral extension of $k[x,y]/(y^2-x^2(x-1))$. 
Also, how does one get to the idea of $t=\frac{y}{x}$? Algebraically if possible, since this is from a commutative algebra class, not algebraic geometry.
 A: Let $R=k[X,Y]/(Y^2-X^2(X-1))$. It is easily shown that $R$ is an integral domain. Moreover, $R$ is not normal since the Jacobian matrix $(-3x^2+2x \;\; 2y)$ has rank zero modulo the maximal ideal $\mathfrak m=(x,y)$, while $2-\dim R_{\mathfrak m}=1$.
Now notice that $t=y/x$ satisfies $t^2=x-1$, so $t$ is integral over $R$. (Now we can provide another reason for $R$ is not integrally closed: $t\notin R$, otherwise $t=f(x,y)$ which is equivalent to $y=xf(x,y)$. By lifting the classes we get $Y-Xf(X,Y)\in(Y^2-X^2(X-1))$ which leads quickly to a contradiction by sending $X$ to $0$.)
Let's show that $R[t]$ is integrally closed. We have $R[t]=k[x,y,y/x]$ and this ring is isomorphic to $$k[X,Y,T]/(Y^2-X^2(X-1),XT-Y,T^2-X+1).$$ In order to show this define a surjective ring homomorphism $\varphi:k[X,Y,T]\to k[x,y,y/x]$ by $X\mapsto x$, $Y\mapsto y$, $T\mapsto y/x$, and note that $(Y^2-X^2(X-1),XT-Y,T^2-X+1)\subseteq\ker\varphi$. Since $\ker\varphi$ is a prime ideal of height one, all we have to do now is to show that $(Y^2-X^2(X-1),XT-Y,T^2-X+1)$ is also a prime ideal. But $$k[X,Y,T]/(Y^2-X^2(X-1),XT-Y,T^2-X+1)\simeq k[T],$$ and thus we get two things: $(Y^2-X^2(X-1),XT-Y,T^2-X+1)=\ker\varphi$ and $R[t]\simeq k[T]$.
From the tower of rings $R\subset R[t]\subset K$ (where $K$ is the field of fractions of $R$) we deduce that the integral closure of $R$ is $R[t]$.
