Question

How do you graph `y^2/100-x^2/75=1` and identify the foci and asympototes?

Answer 1

The foci are F`=(0,5sqrt7)` and F'`=(0,-5sqrt7)`
The asymptotes are `y=(2x)/sqrt3` and `y=-(2x)/sqrt3`

Explanation:

This is an up-down hyperbola.
The general equation is `y^2/a^2-x^2/b^2=1`
The center is `(0,0)`
To determine the foci, we calculate `c=sqrt(a^2+b^2)=+-sqrt175=+-5sqrt7`
The foci are F`=(0,+5sqrt7)` and F'`=(0,-5sqrt7)`

The vertices are `(0,10)` and `(0,-10)`

The slopes of the asymptotes are `+-10/sqrt75=+-10/(5sqrt3)=+-2/sqrt3`
The equations of the asymptotes are `y=(2x)/sqrt3` and `-(2x)/sqrt3`
graph{((y^2/100)-(x^2/75)-1)(y-(2x/sqrt3))(y+(2x/sqrt3))=0 [-74, 74.1, -37.07, 37.06]}