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

1 Answer

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]}