How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola (y-3)^2/25-(x-2)^2/16=1(y−3)225−(x−2)216=1?

1 Answer
Narad T.
Oct 24, 2016

The vertices are (2,8)(2,8) and (2,-2)(2,−2)
The foci are (2,3+sqrt41)(2,3+√41) and (2,3-sqrt41)(2,3−√41)
The equations of the asymptotes are (y=3+5/4(x-3))(y=3+54(x−3))
and (3-5/4(x-3))(3−54(x−3))

Explanation:

Looking at the equation it's an up-down hyperbola

The center of the hyperbola is (2,3)(2,3)

The vertices are (2,8)(2,8) and (2,-2)(2,−2)

The slope of the asymptotes ares (5/4)(54) and (-5/4)(−54)

The equations of the asymptotes are (y=3+5/4(x-3))(y=3+54(x−3))
and (3-5/4(x-3))(3−54(x−3))

To calculate the foci, we need c=+-sqrt(16+25)=sqrt41c=±√16+25=√41

The foci are (2,3+sqrt41)(2,3+√41) and (2,3-sqrt41)(2,3−√41)

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