Question

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

Answer 1

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]}