Question

What is the standard form of this ellipse equation?
4x^2 − 8x + 16y^2 + 64y + 4 = 0
please hurry!

Answer 1

Answer:

$4x^{2} -8x+16y^{2} +64y+4=0\\\\4x^{2} -8x+16y^{2} +64y=-4\\\\-4(-x^{2} +2x-4y^{2}-16y)=-4\\\\-(x^{2} -2x+1-1)-4(y^{2} +4y+4-4)=1\\\\-(x-1)^{2} +1-4(y+2)^{2} +16=1\\\\-(x-1)^{2} -4(y+2)^{2}=1-1-16\\\\-(x-1)^{2} -4(y+2)^{2}=-16\\\\\frac{-(x-1)^{2}}{-16} -\frac{4(y+2)^{2}}{-16} =\frac{-16}{-16} \\\\\frac{(x-1)^{2}}{16} +\frac{(y+2)^{2}}{4} =1$

Answer 2

Answer:

Hello,

Answer :(x-1)²/16+(y+2)²/4=1

Let 's me the time to transfert datas from EU to USA :)

Step-by-step explanation:

$4x^2-8x+16y^2+64y+4=0\\\\4(x^2-2x+1) -4 +16(y^2+4y+4)-64+4=0\\\\4(x-1)^2+16(y+2)^2=64\\\\\\\boxed{\dfrac{(x-1)^2}{16} +\dfrac{(y+2)^2}{4} =1}$