What is the center (x^2)/(36)+(y^2)/(16)=1 ?

The center (x^2)/(36)+(y^2)/(16)=1 is (0,0)

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center (x^2)/(36)+(y^2)/(16)=1

center

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

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{6 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 6, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 6, b = 4

And the center is:

(0, 0)

9 x^{2} - 36 x + 4 y^{2} = 0

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

4 x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{81} = 1

\frac{( x - 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{25} = 1

What is the center (x^2)/(36)+(y^2)/(16)=1 ?
The center (x^2)/(36)+(y^2)/(16)=1 is (0,0)