What is the center (x^2)/9+(y^2)/5 =1 ?

The center (x^2)/9+(y^2)/5 =1 is (0,0)

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center (x^2)/9+(y^2)/5 =1

center

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

Rewrite

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{3 ^{2}} + \frac{( y - 0 ) ^{2}}{( 5 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 3, b = 5

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 3, b = 5

And the center is:

(0, 0)

4 x^{2} + 9 y^{2} - 8 x - 36 y + 4 = 0

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{25} = 1

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

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

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