I've reworked this and added on is any of it correct?Totally confused-learned this today and I don't get it-Please help
Write the standard form of 9x^2 + 4y^2-72x+16y+124=0
I think the first step is to take the constant to right side so I have:
9x^2 + 4y^2-72x+19(x^2-8x+16) + 4(y^2 + 4y + 4) = -124+ 144+16
9(x+4)^2 + 4(y^2 + y) = 36
(x+4)^2/4 + (y^2+y)/9 = 1
5 answers
sure looks like an ellipse in standard form to me
but (y^2 + 4y + 4) is (y+2)^2
9 x^2 -72 x + ?
9 (x^2 - 8 x + 16)
9 (x-4)^2
9 (x^2 - 8 x + 16)
9 (x-4)^2
so
9x^2 + 4y^2-72x+16y+124=0
9 x^2 -72 x + 4 y^2 + 16 y = -124
9(x^2-8x) + 4(y^2+4 y) = -124
9 (x^2-8x+16) + 4(y^2+4y) = -124+9*16
9(x-4)^2 +4(y^2+4y) = 20
9(x-4)^2 + 4(y^2+4y+4) = 20+16
9(x-4)^2+4(y+2)^2 = 36
so
(x-4)^2/4 +(y+2)^2/9 = 1
9x^2 + 4y^2-72x+16y+124=0
9 x^2 -72 x + 4 y^2 + 16 y = -124
9(x^2-8x) + 4(y^2+4 y) = -124
9 (x^2-8x+16) + 4(y^2+4y) = -124+9*16
9(x-4)^2 +4(y^2+4y) = 20
9(x-4)^2 + 4(y^2+4y+4) = 20+16
9(x-4)^2+4(y+2)^2 = 36
so
(x-4)^2/4 +(y+2)^2/9 = 1
ellipse with center at (4,-2)
semi axes of length 2 and 3
semi axes of length 2 and 3