<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] First, you can simply $f(x)$ a little and write
$$ f(x) = \frac{x^2}{x-1} = \frac{x^2-1+1}{x-1} = (x + 1) + \frac{1}{x-1}. $$
Then we can calculate the derivative to obtain:
$$ f'(x) = 1 - \frac{1}{(x-1)^2} = \frac{x^2-2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}. $$
If the line passes through $(2,0)$, it has the form $y = m(x - 2)$. If it also passes through $(a,f(a))$ we must have
$$ f(a) = m(a - 2) \implies m = \frac{f(a)}{(a-2)}. $$
Finally, we must have
$$ f'(a) = m = \frac{f(a)}{(a-2)} \implies \frac{a(a-2)^2}{(a-1)^2} = \frac{a^2}{a - 1} \iff a(a-2)^2 = a^2(a-1). $$
One solution is obtained by taking $a = 0$. For other solutions, we must solve
$$ (a-2)^2 = a^2 - 4a + 4 =  a(a-1) = a^2 - a \implies \\
4 - 3a = 0 \implies a = \frac{4}{3}. $$
</s>