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

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

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

center

\frac{x ^{2}}{9} + y^{2} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{9} + y^{2} = 1

Rewrite

\frac{x ^{2}}{9} + y^{2} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(0, 0)

9 x^{2} + 4 y^{2} + 54 x - 8 y + 49 = 0

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{36} = 1

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

16 x^{2} + 25 y^{2} - 128 x + 150 y + 81 = 0

11 x^{2} + 3 x + 3 y^{2} + 11 y = 0