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#?

1 Answer
Oct 24, 2016

Answer:

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

Explanation:

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

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

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

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

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

To calculate the foci, we need #c=+-sqrt(16+25)=sqrt41#

The foci are #(2,3+sqrt41)# and #(2,3-sqrt41)#