Question 792229
Find location of the center, vertices, and foci.
(x+4)^2/36-(y-1)^2/25=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form of equation:{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}, (h,k)=(x,y) coordinates of center
For given hyperbola:
center:(-4,1)
a^2=36
a=6
b^2=25
b=5
c^2=a^2+b^2=61
c=√61≈7.8
..
vertices: (-4±a,1)=(-4±6,1)=(-10,1) and (2,1)
....foci: (-4±c,1)=(-4±7.8,1)=(-11.8,1) and (3.8,1)