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

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

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

center

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

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(0, 0)

\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

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

y^{2} = 2 (4 x - 3 - x^{2})

6 x^{2} + 9 y^{2} - 18 y - 45 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1