What are the foci of the ellipse
16( y - 3)^2= 16 - (x - 2)^2 ?
These problems are killing me. Test in the morn, still don't get it. First, I divided by 16 to get (y-3)^2 = -(x-2)^2/16
and then convert that to
(y-3)^2 + (x-2)^2/16 = 0 factor out to y2-6y+9 + (x^2-2x+4)/16=0, put the 9 on other side and divide out to get...
y^2/9 - 2y/3 + (x^2-2x+4)/144=1 , then i get a common denominator by multiplying by 3 to get (y^2-6y)/9 +(x^2-2x+4)/144=1
and here i'm stuck... i 'm not sure if i'm making a mistake in the above or not seeing the next step to find a, b
16( y - 3)^2= 16 - (x - 2)^2 ?
These problems are killing me. Test in the morn, still don't get it. First, I divided by 16 to get (y-3)^2 = -(x-2)^2/16
and then convert that to
(y-3)^2 + (x-2)^2/16 = 0 factor out to y2-6y+9 + (x^2-2x+4)/16=0, put the 9 on other side and divide out to get...
y^2/9 - 2y/3 + (x^2-2x+4)/144=1 , then i get a common denominator by multiplying by 3 to get (y^2-6y)/9 +(x^2-2x+4)/144=1
and here i'm stuck... i 'm not sure if i'm making a mistake in the above or not seeing the next step to find a, b