Hi, 

Note: Standard Form of an Equation of an Hyperbola opening up and down is:

   

where Pt(h,k) is a center  with vertices 'b' units up and down from center.

25y^2-4x^2+100y+24x-36=0 

 25[(y+2)^2 -4] -4[(x-3)^2 - 9] - 36 = 0

   25(y+2)^2 -100 - 4(x-3)^2 + 36 - 36 = 0

   25(y+2)^2 - 4(x-3)^2 = 100

  C(3,-2) with vertices V(3,0) and V(3,-4)

Foci: c = sqrt(29)  foci(-3, -2-sqrt(29)) and (-3, -2+sqrt(29))



25y^2-x^2=25

  Yes. C(0,0) vertices(0,1),(0,-1)... foci(0,±sqrt(26))  

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