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

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

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

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

(h, k) = (0, - 1), a = 3, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, - 1), a = 3, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, - 1), b = 2, a = 3

25 x^{2} + 9 y^{2} = 25

x^{2} + 2 y^{2} - 6 x - 20 y + 59 = 12

x^{2} + 4 y^{2} - 46 = 0

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} - 1 = 0

x^{2} + 2 x + 3 y^{2} = 6

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