How do you identify the vertices, foci, and direction of #y^2/25-x^2/16=1#?

1 Answer
Nov 27, 2017

See the explanation below

Explanation:

The equation is

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

#=>#, #(y/5)^2-(x/4)^2=1#

This is the equation of a hyperbola with a vertical transverse axis.

#y^2/a^2-x^2/b^2=1#

The center of the hyperbola is #C=(0,0)#

The vertices are #V= (0,5)# and #V'=(0,-5)#

The foci are #F=(0,c)# and #F'=(0,-c)#

where #c=sqrt(a^2+b^2)=sqrt(25+16)=sqrt41#

The foci are #F=(0,sqrt41)# and #F'=(0,-sqrt41)#

The asymptotes are #y=+-(a/b)x#

The asymptotes are #y=+-(5/4)x#

graph{y^2/25-x^2/16=1 [-32.47, 32.47, -16.25, 16.24]}