SOLUTION: please solve this and graph: 36y^2 + 216y - 4x^2 - 40x +80 = 0 thanks so much.

 

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   

   

Hyperbola opening up and down: C(-5,-3) Vertices (-5,-1) and (-5,-5





Standard Form of an Equation of a Circle is  

where Pt(h,k) is the center and r is the radius



 Standard Form of an Equation of an Ellipse is 

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   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:

   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, 

 where(h,k) is the vertex

 The standard form is , where  the focus is (h,k + p)