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#?

1 Answer
Jun 7, 2018

Answer:

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]}