What is (x^2)/4+((y+3)^2)/9 =1 ?

The solution to (x^2)/4+((y+3)^2)/9 =1 is Ellipse with (h,k)=(0,-3),a=2,b=3

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(x^2)/4+((y+3)^2)/9 =1

\frac{x ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

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

Solution steps

\frac{x ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

Rewrite

\frac{x ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

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

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

\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 11

9 x^{2} + 25 y^{2} + 18 x + 50 y - 191 = 0

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

What is (x^2)/4+((y+3)^2)/9 =1 ?
The solution to (x^2)/4+((y+3)^2)/9 =1 is Ellipse with (h,k)=(0,-3),a=2,b=3