What is the center 2x^2+2y^2+16y=0 ?

The center 2x^2+2y^2+16y=0 is (0,-4)

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center 2x^2+2y^2+16y=0

center

2 x^{2} + 2 y^{2} + 16 y = 0

(0, - 4)

Solution steps

2 x^{2} + 2 y^{2} + 16 y = 0

Rewrite

2 x^{2} + 2 y^{2} + 16 y = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{4 ^{2}} + \frac{( y - ( - 4 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, - 4), a = 4, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, - 4), b = 4, a = 4

And the center is:

(0, - 4)

x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0

13 x^{2} + 49 y^{2} = 637

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{100} = 1