What is the center 0=-400/9 x^2+16y^2+400 ?

The center 0=-400/9 x^2+16y^2+400 is (0,0)

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center 0=-400/9 x^2+16y^2+400

center

0 = - \frac{400}{9} x^{2} + 16 y^{2} + 400

(0, 0)

Solution steps

0 = - \frac{400}{9} x^{2} + 16 y^{2} + 400

Rewrite

0 + \frac{400}{9} x^{2} - 16 y^{2} - 400 = 0

in the form of a standard hyperbola equation

\frac{( x - 0 ) ^{2}}{3 ^{2}} - \frac{( y - 0 ) ^{2}}{5 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, 0), a = 3, b = 5

And the center is:

(0, 0)

x^{2} - y^{2} + 7 x + 8 y = 5

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

9 y^{2} - 49 x^{2} = 441

4 x^{2} - 9 y^{2} + 32 x + 36 y + 64 = 0

4 x^{2} - 45 y^{2} = 180

What is the center 0=-400/9 x^2+16y^2+400 ?
The center 0=-400/9 x^2+16y^2+400 is (0,0)