Find the vertices and asymptotes of the hyperbola y^2+6y-4x^2+5=0

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola.
y^2+6y-4x^2+5=0
Please help out a bit,thank you

Complete the square on the y terms:

y^2 + 6y

= (y + 3)^2 - (3)^2

= (y + 3)^2 - 9

Hence:

y^2 + 6y - 4x^2 + 5 = 0 becomes:

(y + 3)^2 - 9 - 4x^2 + 5 = 0

=> (y + 3)^2 - 4x^2 = 4

Divide through by 4

=> (1/4)(y +3 )^2 - x^2 = 1

=> -x^(2)/1^(2) + (y + 3)^(2)/2^(2) = 1

which is of the form: x^(2)/a^(2) - (y^2)/b^(2) = 1

where a = 1 and b = 2

The eccentricity is given by: e = sqrt(1 + b^2/a^2). Hence e = sqrt(1 + 4/1) = 5

The foci of the hyperbola are at +/- (ae, 0). Hence the foci are at points (5, 0) and (-5,0)

The equations of the asymptotes are y = +/- (b/a)x where a,b >). Hence the equations are:

y = +/-(5x)

Complete the square on the y, (y^2-6y+9)-9-4x^2+5=0
=> (y-3)^2 - 4x^2 =4
=> (y-3)^2/4 - x^2 =1
so a^2=4 and b^1=1
The centre is (0,3)
Vertices are where x=0, giving (y-3)^2=4 => y=1 or 5
so (0,1) and (0,5)
b^2=a^2(e^2-1) giving e=sqrt(5)/2, ae=sqrt(5) and foci (0,3-sqrt(5)) and (0, 3+sqrft(5))
Asymptotes have equations given by (y-3)^/2 - x^2 = 0 giving ????

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