What is the center (x^2)/(81)+(y^2)/(64)=1 ?

The center (x^2)/(81)+(y^2)/(64)=1 is (0,0)

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center (x^2)/(81)+(y^2)/(64)=1

center

\frac{x ^{2}}{81} + \frac{y ^{2}}{64} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{81} + \frac{y ^{2}}{64} = 1

Rewrite

\frac{x ^{2}}{81} + \frac{y ^{2}}{64} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{9 ^{2}} + \frac{( y - 0 ) ^{2}}{8 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 9, b = 8

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 9, b = 8

And the center is:

(0, 0)

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

\frac{( x + 5 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{36} = 1

\frac{( x - 3 ) ^{2}}{86} + \frac{( y + 5 ) ^{2}}{49} = 1

x^{2} + y^{2} - 6 y = 0

4 (x + 5)^{2} + 8 (y - 2)^{2} = 32

What is the center (x^2)/(81)+(y^2)/(64)=1 ?
The center (x^2)/(81)+(y^2)/(64)=1 is (0,0)