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

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

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

center

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

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(0, 0)

6 x^{2} + 8 y^{2} + 12 x - 32 y = 178

8 x^{2} + 2 y^{2} + 16 x = 0

4 x^{2} + 36 y^{2} - 24 x - 288 y + 611 = 0

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

25 x^{2} + 144 y^{2} = 900