What is the center 9x^2+16y^2-18x+96y-279=0 ?

The center 9x^2+16y^2-18x+96y-279=0 is (1,-3)

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

center

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

(1, - 3)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 3), a = 43, b = 33

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, - 3), a = 43, b = 33

And the center is:

(1, - 3)

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

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

What is the center 9x^2+16y^2-18x+96y-279=0 ?
The center 9x^2+16y^2-18x+96y-279=0 is (1,-3)