Question

How do you identify the vertices, foci, and direction of `(y-1)^2/9-(x+1)^2/16=1`?

Answer 1

The vertices are `=(-1,4)` and `=(-1,-2)`
The foci are F`=(-1,6)` and F'`=(-1,-4)`
The equations of the asymptotes are `y=1+3/4(x+1)` and `y=1-3/4(x+1)`

Explanation:

This is the equation of an updown hyperbola.
The standard equation is `(y-k)^2/b^2-(x-h)^2/a^2=1`
The center is `(h,k)=(-1,1)`

The vertices are `(h,k+b)=(-1,4)` and `(h,k-b)=(-1,-2)`

The slopes of the asymptotes are `+-b/a=+-3/4`
The equations of the asymptotes are `y=k+-b/a(x-h)`
`y=1+3/4(x+1)` and `y=1-3/4(x+1)`

To calculate the foci, we need `c=+-sqrt(a^2+b^2)`
So `c=+-sqrt(9+16)=+-sqrt25=+-5`
The foci are `=(h,k+-c)`
so F`=(-1,6)` and F' `(-1,-4)`
graph{((y-1)^2/9-(x+1)^2/16-1)(y-1-3/4(x+1))(y-1+3/4(x+1))=0 [-22.82, 22.77, -11.4, 11.4]}