Question

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`?

Answer 1

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)`