Answer:
(-2, 1) and (6, 1)
Step-by-step explanation:
The standard form equations for a hyperbola are ...
[tex]\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2} = 1 \\\\\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2} = 1[/tex]
The first form opens horizontally; the second opens vertically. Further, the center-focus distance 'c' is given by ...
[tex]c^2 = a^2 +b^2 \qquad\text{$c$ = distance from center to focus}[/tex]
The attached figure illustrates the relation between the various parameters and the features of the hyperbola.
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7.
Using the above information and the information in the first attachment, we find ...
(h, k) = (2, 1)
a = 4, b = 2
The vertices are (h±a, k), so are (2±4, 1) = (-2, 1) and (6, 1).
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The second attachment illustrates the hyperbola and its vertices.
