Find an equation for a tangent line to the curve $x^2- y^2=5$ that passes through the point $(1, 1)$. Find an equation for a tangent line to the curve $$ x^2 - y^2 = 5$$ that passes through the point $(1, 1)$.
I realize that I have to use implicit differentiation $$2x - 2y \frac {dy}{dx} = 0$$ 
$$\frac {dy}{dx} = \frac xy$$
However I do not know how to relate this to the point in order to find the equation of the line that passes through the point, even though the point isn't (presumably) on the curve. 
Thank you in advance for the help.

Here's an approach without (implicit) differentation.
A line passing through $(1,1)$ and with slope $m$ has an equation of the form:
$$y=m(x-1)+1 \iff y = mx-m+1$$
The points of intersection of this line and the hyperbola $x^2-y^2=5$ are the solutions to the following system:
$$\left\{\begin{array}{l}
y = mx-m+1 \\
x^2-y^2=5
\end{array}\right.$$
Substitution of the first into the second equation yields a quadratic equation in $x$:
$$x^2-(mx-m+1)^2=5 \iff \left( 1-m^2 \right)x^2  + \left(2 m^2- 2 m \right) x - m^2  + 2 m - 6 = 0 $$
The line is tangent to the hyperbola if the discriminant is $0$, so:
$$\begin{align}\left(2 m^2- 2 m\right)^2-4\left(1-m^2\right)\left(- m^2  + 2 m - 6 \right)  = 0 & \iff -16 m^2 - 8 m + 24 = 0 \\
& \iff 2m^2 + m -3 = 0  \\
& \iff \left( m-1 \right)\left(2m+3 \right) = 0 
\end{align}$$
This gives you two slopes.
