What is the center 20x^2-16y^2-40x128x-556=0 ?

The center 20x^2-16y^2-40x128x-556=0 is (0,0)

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center 20x^2-16y^2-40x128x-556=0

center

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

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{139}{5 51} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{139}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{139}{5 51}, b = \frac{139}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = \frac{139}{2}, a = \frac{139}{5 51}

And the center is:

(0, 0)

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

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

4 x^{2} + y^{2} + 8 x + 2 y - 31 = 0

\frac{( x + 5 ) ^{2}}{49} + \frac{( y - 2 ) ^{2}}{81} = 1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{36} = 1

What is the center 20x^2-16y^2-40x128x-556=0 ?
The center 20x^2-16y^2-40x128x-556=0 is (0,0)