Question

How do you find the vertices, asymptote, foci and graph `y^2/9-x^2/100=1`?

Answer 1

The vertices are `(0,3)` and `(0,-3)`
The equations of the asymptotes are `y=3/10x` and `y=-3/10x`
the foci are F`(0,sqrt109)` and F'`(0,-sqrt109)`

Explanation:

The general equation is `(y-h)^2/a^2-(x-k)^2/b^2=1`

This is an up down hyperbola and the center is `(0,0)`

The slope of the asympyotes are `+-3/10`
The equations of the asymptotes are `y=3/10x` and `y=-3/10x`
To determine the foci, we need `c=sqrt(a^2+b^2)=+-sqrt109`
`:.` the foci are F`(0,sqrt109)` and F'`(0,-sqrt109)`
graph{(y^2/9)-(x^2/100)=1 [-18.02, 18.01, -9.01, 9.01]}