Question 422335
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Hi
 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.

Finding the center of the following ellipse:  
16x^2+4y^2+64x+24y+36=0
 16(x^2 + 4x) + 4(y^2+6y)+36 = 0   |completing the squares
 16[(x+2)^2 -4] + 4[(y+3)^2-9] + 36 = 0
 16(x+2)^2 - 64 + 4(y+3)^2 - 36 + 36 = 0
   16(x+2)^2 + 4(y+3)^2 = 64
   {{{(x+2)^2/4 + (y+3)^2/16 = 1}}}
center pt(-2,-3)
{{{drawing(300,300,   -10,10,-10,10,  arc(-2,-3,4,8),
 grid(1),
circle(-2, -3,0.4),
graph( 300, 300, -10,10,-10,10))}}}