Finding equation of tangent lines to hyperbola $xy=1$ I am working through some calculus problems (this is in a section on implicit differentiation) and this one is giving me trouble.
I am trying to find the equations of the tangent lines to the hyperbola $$xy=1$$
That pass through the point $(-1,1)$. 
As in the other problems of this type, I implicitly differentiated the relation between $x$ and $y$
$$\frac{dy}{dx}x+y=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$$
Then we have:
$$\frac{dy}{dx}\vert_{(-1,1)}=-(1/-1)=1$$
So I would think lines passing through this point would have the equation:
$$y-1=x+1\Rightarrow y=x+2$$
But this is definitely not true from plotting on wolfram alpha. The solutions in the book are
$$y=\pm(2\sqrt{2}-3)x-2\pm2\sqrt{2} $$
 A: They want the tangent line(s) that pass through the point $(-1,1)$, which is clearly not on the curve.  
Let the point of tangency be $(a,b)$. Then the slope of the tangent line, by your calculation, is $-b/a$. The equation of the tangent line is
$$y-b=(-b/a)(x-a).$$
Since this line passes through $(-1,1)$, we have
$$1-b=(-b/a)(-1-a).$$
Using the fact that $b=1/a$ we get
$$1-\frac{1}{a}=(-1/a^2)(-1-a)$$
This simplifies to $a^2-2a-1=0$. Now we can solve for $a$ and finish.
A: Using homogenous coordinates, your hyperbola can be expressed as the solution set to 
$P(X) = X^TQX = X^T\begin{pmatrix}0&1&0\\1&0&0\\0&0&-2\end{pmatrix}X = 0$
What makes this form neat is that the tangent line at any point of this parabola can be calculated simply by
$QX \equiv Q\begin{pmatrix}x\\y\\1\end{pmatrix}$ where $x,y$ denote the coordinates in the Euclidean plane.
Since the line must also pass through (-1,1) we need only solve 
$\begin{pmatrix}-1&1&1\end{pmatrix}Q\begin{pmatrix}x\\y\\1\end{pmatrix}=0\\
\begin{pmatrix}-1&1&1\end{pmatrix}\begin{pmatrix}y\\x\\-2\end{pmatrix}=0\\$
which gives $x-y-2=0$
We also have the condition $y = 1/x$, combined we have
$x^2-2x-1=0$ which can easily be solved.
