y6 x y2 y4 x y4 y2 x y2 x y4 y2 x y2 x y2 x y2
Yes, as x-y2=0
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
Y3
y2-x-6y+6=0 y2-6y+6=x (y-3)2-3=x so y = 3+sqrt(x+3) or y = 3-sqrt(x+3)
y6 x y2 y4 x y4 y2 x y2 x y4 y2 x y2 x y2 x y2
0
Yes, as x-y2=0
X2+y2=25 (x-8)2+y2 =41
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
x2 - y2 + 9x - 9y =(x2 + 9x) - (y2 + 9y) =x(x + 9) - y(y + 9)================================Another way to go after it:x2 - y2 + 9x - 9y =(x2 - y2) + 9x - 9y =(x + y) (x - y) + 9 (x - y) =(x + y + 9) (x - y)
x2-y2=(x-y)(x+y) which is a well known identity.
4 - y2 can be written as 22 - y2. This has become of the form of x2 - y2.Expansion for x2 - y2 is (x+y)(x-y).So, 4 - y2 = 22 - y2 = (2+y)(2-y)
(x - y)(x + y)(x2 - xy + y2)(x2 + xy + y2)
-2
(x-2)^2+y^2=64
X = y2 =====