Example 1 Sketch the graph of each of the following hyperbolas.
Solution
(a) 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 (3,−1) and as always watch the signs! Once we have the center we
can get the vertices. These are (8,−1) and (−2,−1).
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
.
.
Now that we’ve got the center and the slopes of the asymptotes we
can get the equations for the asymptotes. They are,
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) 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 (−2,0) .
Remember that since there is a y2 term by itself we had to have k = 0 .
At this point we also know that the vertices are (−2,3) and (−2,−3) .
In order to see
the slopes of the asymptotes let’s rewrite the equation a little.
So, the slopes of the asymptotes are. The equations of the
asymptotes are then,
y = 0 + 3(x + 2) = 3x
+ 6 and y = 0 − 3(x + 2) = −3x
− 6
Here is the
sketch of this hyperbola.
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