Question 448875
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Hi
36y^2 + 216y - 4x^2 - 40x +80 = 0
[36(y+3)^2 - 9] -4[(x+5)^2 -25] = -80   |Multiplying thru by -1
  36(y+3)^2 - 324 -4(x+5)^2 + 100 = -80
  36(y+3)^2 -4(x+5)^2 = 144   
   {{{(y+3)^2/4 -(x+5)^2/36 = 1 }}}
Hyperbola opening up and down: C(-5,-3) Vertices (-5,-1) and (-5,-5
{{{drawing(300,300,  -10,10,-10,10, grid(1), 
circle(-5, -1,0.4),
circle(-5, -5,0.4),

graph( 300, 300,-10,10,-10,10,2sqrt((x+5)^2/36 + 1)-3,-2sqrt((x+5)^2/36 +1)-3))}}}

Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}} 
where Pt(h,k) is the center and r is the radius

 Standard Form of an Equation of an Ellipse is {{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1 }}}
where Pt(h,k) is the center and a and b  are the respective vertices distances from center.

Standard Form of an Equation of an Hyperbola is  {{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}} where Pt(h,k) is a center  with vertices 'a' units right and left of center.
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.

Using the vertex form of a parabola opening up or down, {{{y=a(x-h)^2 +k}}}
 where(h,k) is the vertex
 The standard form is {{{(x -h)^2 = 4p(y -k)}}}, where  the focus is (h,k + p)