Question 451429
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Hi, 
Note: Standard Form of an Equation of an Hyperbola opening up and down is:
  {{{(y-k)^2/b^2 - (x-h)^2/a^2 = 1}}} 
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
{{{(y+2)^2/4 -(x-3)^2/25 =1}}}  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))
{{{drawing(300,300,-10,10,-10,10,  grid(1),
circle(3, -2,0.3),
circle(3, 0,0.3),
circle(3, -4,0.3),
circle(3, 3.4,0.3),
circle(3, 3.4,0.3),
circle(3, -7.4,0.3),
graph(300,300,-10,10,-10,10,0,2sqrt(1+ (1/25)(x-3)^2)-2,-2sqrt(1+ (1/25)(x-3)^2)-2))}}}
25y^2-x^2=25
 {{{ y^2/1 - x^2/25 = 1}}} Yes. C(0,0) vertices(0,1),(0,-1)... foci(0,ħsqrt(26))