Question

How do you find all the critical points to graph `(x+3)^2/25-(y+5)^2/4=1` including vertices, foci and asymptotes?

Answer 1

The vertices are `(2,-5)` and `(-8,-5)`
The equations of the asymptotes are `y=-5+2/5(x+3)` and `y=-5-2/5(x+3)`
The foci are `(-3+sqrt29,-5)` and `(-3-sqrt29,-5)`

Explanation:

Our equation is `(x+3)^2/25-(y+5)^2/4=1`
This is the equation of a left to right hyperbola
The typical equation of a hyperbola is `(x-h)^2/a^2-(y-k)^2/b^2=1`
The center is `(h,k)=(-3,-5)`
The vertices are `(h+a,k)=(-3+5,-5)=(2,-5)` and `(h-a,k)=(-3-5,-5)=(-8,-5)`
The slope of the asymptotes are `b/a=2/5` and `-b/a=-2/5`

The equation of the asymptotes are `y=k+b/a(x-h)` and `y-b/a(x-h)`

So the equations are `y=-5+2/5(x+3)` and `y=-5-2/5(x+3)`

In order to calculate the foci, we need `c=+-sqrt(a^2+b^2)=+-sqrt(25+4)=+-sqrt29`
And the foci are `(-3+sqrt29,-5)` and `(-3-sqrt29,-5)`

graph{(x+3)^2/25-(y+5)^2/4=1 [-20.91, 18.56, -13.78, 5.97]}