What is the center 36x^2+9y^2+36y-288=0 ?

The center 36x^2+9y^2+36y-288=0 is (0,-2)

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center 36x^2+9y^2+36y-288=0

center

36 x^{2} + 9 y^{2} + 36 y - 288 = 0

(0, - 2)

Solution steps

36 x^{2} + 9 y^{2} + 36 y - 288 = 0

Rewrite

36 x^{2} + 9 y^{2} + 36 y - 288 = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{3 ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, - 2), a = 3, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, - 2), b = 6, a = 3

And the center is:

(0, - 2)

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{16} = 1

x^{2} + 7 x + 3 y^{2} + y = 0

9 x^{2} + 4 y^{2} = 36

9 X^{2} + 4 Y^{2} = 36

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

What is the center 36x^2+9y^2+36y-288=0 ?
The center 36x^2+9y^2+36y-288=0 is (0,-2)