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Find the center of this ellipse: 4x^2 + 16y^2 +64x +64y + 256 =0 ANSWER PLEASE
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4(x+8)^2+16(y+2)^2=-256+256+64 ((x+8)/4)^2 + ((y+2)/2)^2 = 1 Can you figure out the centre?
No i can't and i reall would appreciate the answer..
An ellipse centred at origin has the equation (x/a)^2 + (y/b)^2 = 1 If centred at (x0, y0), the equation becomes: ((x-x0)/a)^2 + ((y-y0)/b)^2 =1 By comparison with the equation ((x+8)/4)^2 + ((y+2)/2)^2 = 1 can you find x0 and y0?
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