Finding a tangent of an algebraic curve: $xy = 1$ [Well written explanation] I want to find, using (easy) calculus, the slope of a tangent to the algebraic curve $xy = 1$.


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*The tangent I want to find is the tangent that passes through the point $(x_i,y_i) = (0,f(t))$.

*$f$ is defined as $f(t) = 4\cdot2e^{-t/10}$.
I know the general way of finding the slope of a tangent to a function that passes through a point $P$ outside of the function is to put:
$$y-y_i = y'(x-x_i)$$

The problem for me is then to find $x_i$. I've tried the following:
Implicit differentiation of $xy=1$ gives me: 
$$y' = -\frac{y}{x}$$
I then attempt to use the general method of finding tangents that pass through a point.


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*$y-0 = -\dfrac{y}{x}\left(x-4\cdot2^{t/10}\right) \iff 1 = -1+\dfrac{1}{x}\cdot4\cdot2^{t/10} \iff x=2\cdot2^{t/10}$


I'm not sure if I even chose the right approach. Any help appreciated, thank you in advance.

 A: It doesn't matter whether we differentiate implicitly or explicitly. Explicit is not difficult in this case. We have $y=\frac{1}{x}$ so $\frac{dy}{dx}=-\frac{1}{x^2}$.
Let $\left(a,\frac{1}{a}\right)$ be the point of tangency. The tangent line passes through  $(0,f(t))$ so it has slope
$$\frac{\frac{1}{a}-f(t)}{a-0}.$$
The tangent line has slope $-\frac{1}{a^2}$ and therefore
$$\frac{\frac{1}{a}-f(t)}{a-0}=-\frac{1}{a^2}.\tag{1}$$
We solve for $a$, by multiplying both sides by $a^2$. So we get $1-af(t)=-1$and therefore $a=\frac{2}{f(t)}$.
Now that we know the point of tangency, we can find the equation of the tangent line in the usual way. 
Remarks: $1.$ Implicit differentiation makes things marginally simpler-looking. For that approach we would let the point of tangency be $(a,b)$. I think that for most students seeing these things for the first time, the approach we took is more natural.
$2.$ An alternative is to note that the tangent line at $(a,b)$ has equation
$$y-b=-\frac{1}{a^2}(x-a).$$
When $x=0$, we have $y=f(t)$. Substitute in the above equation, and note that $b=\frac{1}{a}$.  We can then solve for $a$ in terms of $f(t)$, and write down the equation of the tangent line. 
$3.$ Note that we called the point of tangency $(a,1/a)$ or $(a,b)$. Calling it $(x,1/x)$ or $(x,y)$ is not a good idea, since we risk confusion with the $x,y$ in the equation of the tangent line.
