What is 3(y^2+x^2-4y)=0 ?

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

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3(y^2+x^2-4y)=0

3 (y^{2} + x^{2} - 4 y) = 0

(h, k) = (0, 2), a = 2, b = 2

Solution steps

3 (y^{2} + x^{2} - 4 y) = 0

Rewrite

3 (y^{2} + x^{2} - 4 y) = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 2), a = 2, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 2), b = 2, a = 2

16 x^{2} + 16 y^{2} - 8 y^{2} - 15 = 0

x^{(2)} + 4 y^{(2)} + 7 = 12 y - 2 x

\frac{( x - 4 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{36} = 1

\frac{( x - 1 ) ^{2}}{4 ^{2}} + \frac{( y - 5 ) ^{2}}{b ^{2}} = 1

x^{(2)} + 3 y^{(2)} - 6 x + 6 y = 0

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