Pre cal ugh!!!!

Mj12

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May 2, 2010
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Find the foci and vertices of the hyperbola. (x-2)^2/16-(y+1)^2/9=1 I do not have my graphing calculator and/or formulas, so if you could add formulas it would be much appreciated. Thank you.- Megan Breanne
 
The foci are a distance c from the origin.

The vertices are at V(-2,-1) and at V(6,-1)

\(\displaystyle c^{2}=a^{2}+b^{2}\Rightarrow \sqrt{4^{2}+3^{2}}=5\)

Since the center is at (2,-1), then the foci are at 2+5=7, (7,-1)

and at 2-5=-3, (-3,-1).
 
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