What is the center (x^2}{36}+\frac{y^2)/9 =1 ?

The center (x^2}{36}+\frac{y^2)/9 =1 is (0,0)

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center (x^2}{36}+\frac{y^2)/9 =1

center

\frac{x ^{2}}{36} + \frac{y ^{2}}{9} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{36} + \frac{y ^{2}}{9} = 1

Rewrite

\frac{x ^{2}}{36} + \frac{y ^{2}}{9} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{6 ^{2}} + \frac{( y - 0 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 6, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 6, b = 3

And the center is:

(0, 0)

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

25 (x + 1)^{2} + 9 (y - 2)^{2} = 225

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

\frac{( x - 6 ) ^{2}}{5} + \frac{( y + 1 ) ^{2}}{6} = 1

36 x^{2} + 5 y^{2} - 90 y - 495 = 0

What is the center (x^2}{36}+\frac{y^2)/9 =1 ?
The center (x^2}{36}+\frac{y^2)/9 =1 is (0,0)