Question

How do you identify the vertices, foci, and direction of `(x-1)^2/4-(y-3)^2/4=1`?

Answer 1

The vertices are `(3,3)` and `(-1,3)`
The foci are `F(1+2sqrt2,3)` and `F'(1-2sqrt2,3)`
The asymptotes are `y=x+2` and `y=-x+4`

Explanation:

This ia a left right hyperbola.
`(x-h)^2/a^2-(y-k)^2/b^2=1`
The center is `(h,k)=(1,3)`
The vertices are `(h+a,k)=(3,3)` and `(h-a,k)=(-1,3)`
The slope of the asymptotes are `+-b/a=+-1`
The equations of the asymptotes are `y=k+-b/a(x-h)`
`=>` `y=3+1(x-1)=x+2`
and `y=3-(x-1)=-x+4`
To calculate the foci, we need `c=+-sqrt(a^2+b^2)`
`=>` `c=+-sqrt8=+-2sqrt2`
the foci are `h+-c,k`
`=>` `F(1+2sqrt2,3)` and `F'(1-2sqrt2,3)`
graph{((x-1)^2/4-(y-3)^2/4-1)(y-x-2)(y+x-4)=0 [-10, 10, -2, 8]}