<s>[INST] Calculate equation of lines tangent to $\frac{x^2}{x-1}$ but also going through $(2,0)$ As the question states, I want to calculate the equations of two lines tangent to $\frac{x^2}{x-1}$, but also going through point $(2,0)$. Here's what I did:

Suppose there is some point at which the line touches the curve, $a$, which will obviously obey the equation $\left(a,\frac{a^2}{a-1}\right)$. The derivative is calculated as
\begin{align*}
f(x) = \frac{x^2}{x-1} = x^2(x-1)^{-1} \Rightarrow f'(x) = 2x(x-1)^{-1} -x^2(x-1)^{-2} = \frac{x^2-2}{(x-1)^2} \Rightarrow f'(a) = \frac{a^2-2}{(a-1)^2}.
\end{align*}
Filling this into the equation we get
\begin{align*}
y - \left(\frac{a^2}{a-1}\right) = \frac{a^2-2}{(a-1)^2}\left(x-a\right).
\end{align*}
Obviously this equation will contain the point $(2,0)$, so we fill it in and obtain the answer:
\begin{align*}
\frac{a^2}{a-1} = \frac{a^2-2}{(a-1)^2}\left(a-2\right) \Rightarrow a^2 +2a-4=0 \Rightarrow (a+1)^2=5 \Rightarrow a = -1\pm\sqrt 5.
\end{align*}

The equations are getting so complicated I'm pretty sure I'm doing it wrong, but I'm not quite sure what I'm doing wrong so I'm wondering if StackExchange could offer some input on my method?
 [/INST] There's a simpler way to solve: take a line through $(2,0)$ with variable slope $t$:
$$y=t(x-2),$$
and write the equation for the intersection points with the curvbe has a double root:
$$t(x-2)=\frac{x^2}{x-1}\iff x^2=t(x-1)(x-2)\iff (1-t)x^2+3tx-2t=0.$$
There's a double root if and only if
$$\Delta=9t^2+8t(1-t)=t(t+8)=0,\enspace \text{i.e.}\quad  t=0,\,-8.$$
Furthermore, the double root is equal to $\;x=\dfrac{3t}{2(t-1)}=0,\,\dfrac 43.$ 
</s>