What is the center 8x^2+2y^2+16x=0 ?

The center 8x^2+2y^2+16x=0 is (-1,0)

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center 8x^2+2y^2+16x=0

center

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

(- 1, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 0), a = 1, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 0), b = 2, a = 1

And the center is:

(- 1, 0)

4 x^{2} + 36 y^{2} - 24 x - 288 y + 611 = 0

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{4} = 1

25 x^{2} + 144 y^{2} = 900

x^{2} + 4 \cdot y^{2} - 4 \cdot y = 0

36 x^{2} + 288 x + 64 y^{2} - 256 y = 1472