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

The center (x^2)/4+((y+3)^2)/9 =1 is (0,-3)

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

center

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

(0, - 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

And the center is:

(0, - 3)

9 x^{2} + 4 y^{2} + 18 x - 8 y - 131 = 0

x^{2} + 4 y^{2} = 4

x^{2} + 4 y^{2} = 1

\frac{( x - 2 ) ^{2}}{36} + \frac{( y - 6 ) ^{2}}{4} = 1

\frac{( x - 5 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{4} = 1

What is the center (x^2)/4+((y+3)^2)/9 =1 ?
The center (x^2)/4+((y+3)^2)/9 =1 is (0,-3)