For (x-4)^2/36 + (y-1)^2/25 = 1, are the vertices (-2, 1) and (10, 1)?
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:%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=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)
that's the answer