What is the center 5y^2-16x^2+400=0 ?

The center 5y^2-16x^2+400=0 is (0,0)

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center 5y^2-16x^2+400=0

center

5 y^{2} - 16 x^{2} + 400 = 0

(0, 0)

Solution steps

5 y^{2} - 16 x^{2} + 400 = 0

Rewrite

5 y^{2} - 16 x^{2} + 400 = 0

in the form of a standard hyperbola equation

\frac{( x - 0 ) ^{2}}{5 ^{2}} - \frac{( y - 0 ) ^{2}}{( 4 5 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, 0), a = 5, b = 45

And the center is:

(0, 0)

4 x^{2} - 9 y^{2} - 16 x + 18 y - 7 = 0

x^{2} - y^{2} + 6 x + 10 y = 2

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

y^{2} - x^{2} = \frac{1}{25}

\frac{1}{144} x^{2} - \frac{1}{196} y^{2} = 1