identifying properties of a hyperbola*?

Question

i thought it was an equation of a circle but turns out its a hyperbola! =)

9x^2-25(y-1)^2=225

so can anyone help me determine the center, foci and vertices from the equation?

Answer 1

The equation of a hyperbola with center at (h.k) is:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
In your equation, h=0, k=1
Divide both sides of equation by 225 to get:
x^2/(9/225) -( y-1)^2/(25/225) = 1
x^2/(3/15)^2 - (y-1)^2/(5/15)^2 = 1
Thus a =3/15=1/5 and b= 5/15=1/3
So the center of the hyperbola is (0,1)
The vertices are at (-a,1) and (a,1) or (-1/5,1) and (1/5,1)
c^2=b^2+a^2= (1/3)^2 + (1/5)^2 = 1/25 + 1/9 = 34/225
So c= sqrt(34)/15
So foci are at (-sqrt(34)/15, 1) and (sqrt(34)/15,1)

Answer 2

to write itin the standard form
x^2/(225/9)-(y-1)^2/(225/25)=1
x^2/25-(y-1)^2/9=1
now you can calculate by comparing it to the standard form
(x^2/a^2-y^2/b^2=1
here a=5 and b=3