Tuesday

March 31, 2015

March 31, 2015

Posted by **Anonymous** on Monday, March 24, 2014 at 9:45am.

- Math - PreCalc (12th Grade) -
**Steve**, Monday, March 24, 2014 at 11:17amthere are lots of them, depending on the eccentricity. Recall that an hyperbola with equation

y^2/a^2 - x^2/b^2 = 1

has vertices at (0,±a) and foci at (0,±c) where c = ae and e>1 is the eccentricity.

Also, recall that c^2 = a^2 + b^2

Here, we have the center of the graph at (2,2) with c=5. So, we can pick a and b as long as a^2+b^2 = 5^2. A happy relation, no? How about a=3,b=4?

So, your hyperbola is shifted some, making its equation

(y-2)^2/9 - (x-2)^2/16 = 1

Verify this at

http://www.wolframalpha.com/input/?i=plane+curve+%28y-2%29^2%2F9+-+%28x-2%29^2%2F16+%3D+1

- Math - PreCalc (12th Grade) -
**Steve**, Monday, March 24, 2014 at 11:18amor, to see the foci plotted,

http://www.wolframalpha.com/input/?i=foci+of+%28y-2%29^2%2F9+-+%28x-2%29^2%2F16+%3D+1

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