What is x^2+(3y-3)^2=4 ?

The solution to x^2+(3y-3)^2=4 is Ellipse with (h,k)=(0,1),a=2,b= 2/3

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x^2+(3y-3)^2=4

x^{2} + (3 y - 3)^{2} = 4

(h, k) = (0, 1), a = 2, b = \frac{2}{3}

Solution steps

x^{2} + (3 y - 3)^{2} = 4

Rewrite

x^{2} + (3 y - 3)^{2} = 4

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{2 ^{2}} + \frac{( y - 1 ) ^{2}}{( \frac{2}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 1), a = 2, b = \frac{2}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 1), a = 2, b = \frac{2}{3}

(x - 5)^{2} + 2 (y - 6)^{2} = 36

9 y^{2} = 144 - 16 x^{2}

x^{2} + 4 y^{2} + 10 x + 9 y - 7 = 0

16 x^{2} + 9 y^{2} - 32 x + 90 y + 145 = 0

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

What is x^2+(3y-3)^2=4 ?
The solution to x^2+(3y-3)^2=4 is Ellipse with (h,k)=(0,1),a=2,b= 2/3