What is the center (x^2)/(64)+(y^2)/(100)=1 ?

The center (x^2)/(64)+(y^2)/(100)=1 is (0,0)

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center (x^2)/(64)+(y^2)/(100)=1

center

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1

Rewrite

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 8, b = 10

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 10, a = 8

And the center is:

(0, 0)

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

9 x^{2} + 16 y^{2} - 54 x + 64 y + 1 = 0

9 x^{2} + 16 y^{2} - 72 x + 64 y - 368 = 0

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

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

What is the center (x^2)/(64)+(y^2)/(100)=1 ?
The center (x^2)/(64)+(y^2)/(100)=1 is (0,0)