What is the center 9x^2+16y^2+126x+96y+441=0 ?

The center 9x^2+16y^2+126x+96y+441=0 is (-7,-3)

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

center

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

(- 7, - 3)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - ( - 7 ) ) ^{2}}{4 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 7, - 3), a = 4, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 7, - 3), a = 4, b = 3

And the center is:

(- 7, - 3)

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

\frac{x ^{2}}{4} + y^{2} = 1

5 x^{2} + 9 y^{2} - 10 x + 4 = 0

What is the center 9x^2+16y^2+126x+96y+441=0 ?
The center 9x^2+16y^2+126x+96y+441=0 is (-7,-3)