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Section 4.4 : Hyperbolas

1. Sketch the graph of the following hyperbola.

\[\frac{{{y^2}}}{{16}} - \frac{{{{\left( {x - 2} \right)}^2}}}{9} = 1\]

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The first step here is to simply compare our equation to the standard form of the hyperbola and identify all the important information. For reference purposes here is the standard form of the hyperbola that matches the one we have here.

\[\frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} = 1\]

Comparing our equation to this we can see we have the following information.

\[h = 2\hspace{0.25in}k = 0\hspace{0.25in}a = 3\hspace{0.25in}b = 4\]

Because the \(y\) term is the positive term we know that this hyperbola will open up and down.

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With the information we found in the first step we can see that the center of the hyperbola is \(\left( {2,0} \right)\).

The vertices of hyperbola are : \(\left( {2, - 4} \right)\) and \(\left( {2,4} \right)\).

The equations of the two asymptotes are,

\[y = 0 + \frac{4}{3}\left( {x - 2} \right) = \frac{4}{3}x - \frac{8}{3}\hspace{0.25in}y = 0 - \frac{4}{3}\left( {x - 2} \right) = - \frac{4}{3}x + \frac{8}{3}\]
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Here is a sketch of the hyperbola including the points and asymptotes we found above.

The graph of the hyperbola \(\frac{{{y^2}}}{{16}} - \frac{{{{\left( {x - 2} \right)}^2}}}{9} = 1\).  It is centered at \(\left( {2,0} \right)\) and opens up and down with vertices at \(\left( {2,4} \right)\) and \(\left( {2, - 4} \right)\), both of which are marked and labeled.  The two blue dashed asymptotes cross at the center and are labeled \(y = \frac{4}{3}x - \frac{8}{3}\) and \(y = - \frac{4}{3}x + \frac{8}{3}\), and the two branches of the hyperbola get closer and closer to them as you move away from the center.