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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?

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2 Answers 2

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.


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To add to Kaster's answer, there is a handy construction elaborated in this link.

In the form


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.

Drawing the hyperbola

Now, in the form


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.

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