What is the center 16x^2+25y^2-128x+150y+81=0 ?

The center 16x^2+25y^2-128x+150y+81=0 is (4,-3)

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center 16x^2+25y^2-128x+150y+81=0

center

16 x^{2} + 25 y^{2} - 128 x + 150 y + 81 = 0

(4, - 3)

Solution steps

16 x^{2} + 25 y^{2} - 128 x + 150 y + 81 = 0

Rewrite

16 x^{2} + 25 y^{2} - 128 x + 150 y + 81 = 0

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{5 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, - 3), a = 5, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (4, - 3), a = 5, b = 4

And the center is:

(4, - 3)

11 x^{2} + 3 x + 3 y^{2} + 11 y = 0

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

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

What is the center 16x^2+25y^2-128x+150y+81=0 ?
The center 16x^2+25y^2-128x+150y+81=0 is (4,-3)