What is the center 16x^2+81y^2=81 ?

The center 16x^2+81y^2=81 is (0,0)

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center 16x^2+81y^2=81

center

16 x^{2} + 81 y^{2} = 81

(0, 0)

Solution steps

16 x^{2} + 81 y^{2} = 81

Rewrite

16 x^{2} + 81 y^{2} = 81

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{9}{4} ) ^{2}} + \frac{( y - 0 ) ^{2}}{1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{9}{4}, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = \frac{9}{4}, b = 1

And the center is:

(0, 0)

9 x^{2} + 16 y^{2} - 18 x + 96 y - 279 = 0

4 x^{2} + 9 y^{2} - 24 x - 36 y = - 36

20 x^{2} - 16 y^{2} - 40 x 128 x - 556 = 0

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

9 x^{2} + 16 y^{2} + 126 x + 96 y + 441 = 0