a \(\displaystyle \frac{{{{\left( {x - 3} \right)}^2}}}{{25}} - \frac{{{{\left( {y + 1} \right)}^2}}}{{49}} = 1\)

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Now, notice that the \(y\) term has the minus sign and so we know that we’re in the first column of the table above and that the hyperbola will be opening left and right.

The first thing that we should get is the center since pretty much everything else is built around that. The center in this case is \(\left( {3, - 1} \right)\) and as always watch the signs! Once we have the center we can get the vertices. These are \(\left( {8, - 1} \right)\) and \(\left( { - 2, - 1} \right)\).

Next, we should get the slopes of the asymptotes. These are always the square root of the number under the \(y\) term divided by the square root of the number under the \(x\) term and there will always be a positive and a negative slope. The slopes are then \( \pm \frac{7}{5}\).

Now that we’ve got the center and the slopes of the asymptotes we can get the equations for the asymptotes. They are,

\[y = - 1 + \frac{7}{5}\left( {x - 3} \right)\hspace{0.25in}{\mbox{and}}\hspace{0.25in}y = - 1 - \frac{7}{5}\left( {x - 3} \right)\]

We can now start the sketching. We start by sketching the asymptotes and the vertices. Once these are done we know what the basic shape should look like so we sketch it in making sure that as \(x\) gets large we move in closer and closer to the asymptotes.

Here is the sketch for this hyperbola.

b \(\displaystyle \frac{{{y^2}}}{9} - {\left( {x + 2} \right)^2} = 1\)

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In this case the hyperbola will open up and down since the \(x\) term has the minus sign. Now, the center of this hyperbola is \(\left( { - 2,0} \right)\). Remember that since there is a *y*^{2} term by itself we had to have \(k = 0\). At this point we also know that the vertices are \(\left( { - 2,3} \right)\) and \(\left( { - 2, - 3} \right)\).

In order to see the slopes of the asymptotes let’s rewrite the equation a little.

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

So, the slopes of the asymptotes are \( \pm \frac{3}{1} = \pm 3\). The equations of the asymptotes are then,

\[y = 0 + 3\left( {x + 2} \right) = 3x + 6\hspace{0.25in}{\mbox{and }}\hspace{0.25in}\,\,y = 0 - 3\left( {x + 2} \right) = - 3x - 6\]

Here is the sketch of this hyperbola.