Calculate equation of lines tangent to $\frac{x^2}{x-1}$ but also going through $(2,0)$

Question

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:

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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*}

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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?

Answer 1

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 $\;x=\dfrac{3t}{2(t-1)}=0,\,\dfrac 43.$

Answer 2

Your method is correct, but note that, as noted in the comment, your derivative is wrong.

In general: if a line passes thorough the points $P=(x,f(x))$ and $Q=(x_Q,y_Q)$ than its slope is : $$ m=\frac{f(x)-y_Q}{x-x_Q} $$

and if the line is tangent to the function $y=f(x)$ at $P$ than this slope is the value of the derivative at this point: $m=f'(x)$, so we have the equation: $$ \frac{f(x)-y_Q}{x-x_Q}=f'(x) $$ where $x$ is the abscissa of the point of tangency.

Answer 3

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

$$ (a-2)^2 = a^2 - 4a + 4 = a(a-1) = a^2 - a \implies \\ 4 - 3a = 0 \implies a = \frac{4}{3}. $$