Graphing Hyperbolas
Sophia Terry 
I know that a Hyperbola is in the form of: $$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$$ But how would I graph it? I know that a Hyperbola has two asymptotes that the graph gets infinitely close to but will never touch, is there a way to find the asymptotes with that equation? and is the asymptotes the only thing you need to graph a hyperbola?
$\endgroup$3 Answers
$\begingroup$To add to Kaster's answer, there is a handy construction elaborated in this link.
In the form
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
the hyperbola has its fundamental rectangle with corners at $(0,b), (0,-b), (a,0), (-a,0)$ and the diagonals of this rectangle are the asymptotes. Points $(a,0), (-a,0)$ are the vertices. Knowing the asymptotes and the vertices, the hyperbola is defined unambiguously. If you want to have a more precise graph when drawing by hand, you may want to calculate additional points for the hyperbola, though.
Now, in the form
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
this whole construction is simply shifted $h$ units in positive $x$ direction and $k$ units in positive $y$ direction.
The most general form
$A_{xx}x^2 + 2A_{xy}xy + A_{yy}y^2 + 2B_x x + 2B_y y + C = 0$
is a bit more problematic, though, since the fundamental triangle is rotated.
$\endgroup$ $\begingroup$First, do a handy parametrization of the hyperbola. To do that, recall the Pythagorean theorem for hyperbolic functions $$ \cosh^2t-\sinh^2t=1 $$ So, you can substitute $$ \frac {(x-h)^2}{a^2} = \cosh^2 t \\ \frac {(y-k)^2}{b^2} = \sinh^2 t $$ which leads to the parametrization $$ x = h \pm a \cosh t \\ y = k + b \sinh t $$ Here $x$-parametrization has $\pm$ sign due to the even nature of $\cosh t$ function, that's the only way to cover negative values of $x-h$.
Due to the symmetry, let's consider only one half of the hyperbola. $$ x = h + a \cosh t \\ y = k + b \sinh t $$ To find its asymptote one can check existence of the limit $$ p = \lim_{x \to \infty} \frac yx = \lim_{t \to \pm \infty} \frac {k + b \sinh t}{h + a \cosh t} = \lim_{t \to \pm \infty} \frac {\frac k{\cosh t} + b \tanh t}{\frac h{\cosh t}+a} = \pm \frac ba $$ So, this limit exists and so do the asymptote(s). To find where these lines intersect $Oy$-line, find the following limit(s) $$ q = \lim_{x \to \infty} (y - px) = \lim_{t \to \pm \infty} \left [k + b \sinh t \mp \frac ba \left (h + a \cosh t \right )\right ] = \\ = k \mp \frac ba h \pm b\lim_{t \to \pm \infty} \left ( \sinh |t| - \cosh |t|\right ) = k \mp \frac ba h \pm b\lim_{t \to \pm \infty} \left(\frac {e^{|t|}-e^{-|t|}}2-\frac {e^{|t|}+e^{-|t|}}2 \right ) = \\ = k \mp \frac ba h \mp b\lim_{t \to \pm \infty} e^{-|t|} = k \mp \frac ba h $$ Remember to take corresponding sign in all expressions simultaneously. So asymptotes are $$ y = px+q = \pm \frac ba x + k \mp \frac ba h = \pm \frac ba (x - h) + k $$ By similar analysis for the second half of the parabola you may find that the same lines are asymptotes too.
To solidify, let's take some example. $$ \frac {(x-2)^2}{3^2}-\frac {(y+2)^2}{4^2}=1 $$ so, $a = 3$, $b = 4$, $h = 2$ and $k = -2$. $$x = 2 \pm 3 \cosh t \\ y = -2 + 4 \sinh t $$ and asymptotes $$ y = \pm \frac 43 (x-2)-2 $$ Schematics is below.
As you have said, the equation of a hyperbola is $$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$$ The distance between the vertices is $2a$. The distance between the foci is $2c$.
A hyperbola opening left/right (has a horizontal transverse axis) and center at $(h, k)$ has one asymptote with equation $y = k + \frac{b}{a}(x - h)$ and the other with equation $y = k - \frac{b}{a}(x - h)$. A hyperbola with a vertical transverse axis and center at $(h, k)$ has one asymptote with equation $y = k + \frac{a}{b}(x - h)$ and the other with equation $y = k - \frac{a}{b}(x - h)$.
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