What is the center (x^2)/9+(y^2)/(16)>1 ?

The center (x^2)/9+(y^2)/(16)>1 is (0,0)

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center (x^2)/9+(y^2)/(16)>1

center

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

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{3 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 4, a = 3

And the center is:

(0, 0)

4 x^{2} + 9 y^{2} = 36

25 x^{2} + 9 y^{2} = 225

4 x^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

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

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

What is the center (x^2)/9+(y^2)/(16)>1 ?
The center (x^2)/9+(y^2)/(16)>1 is (0,0)