Question

How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola `y^2/16-x^2/25=1`?

Answer 1

Vertices are at `(0,4) and (0,-4)` , Focii are at
`(0,6.4) and (0,-6.4)` Asymptotes are `y= +- 4/5 x`

Explanation:

`y^2/16-x^2/25=1`

The standard equation of vertical hyperbola is

`(y-k)^2/a^2-(x-h)^2/b^2=1:. h=0, k=0 , a=4 , b=5`

Center isat `0,0` Vertices are at `(0,a) and (0,-a)` or

Vertices are at `(0,4) and (0,-4)`

`c^2=a^2+b^2 = 4^2+5^2=41 :. c= +-sqrt41~~ +-6.4`

Foci are `c` units from the center. Therefore,

Focii are at `(0,6.4) and (0,-6.4)` equation of asymptotes

of vertical hyperbola are `y-k =+- a/b(x-h)`

Asymptotes are : `y-0= +- 4/5( x-0) or y= +- 4/5 x`

graph{y^2/16-x^2/25=1 [-12.67, 12.64, -6.33, 6.33]}