What is the center x^2-8x+8y^2+12y=0 ?

The center x^2-8x+8y^2+12y=0 is (4,-3/4)

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center x^2-8x+8y^2+12y=0

center

x^{2} - 8 x + 8 y^{2} + 12 y = 0

(4, - \frac{3}{4})

Solution steps

x^{2} - 8 x + 8 y^{2} + 12 y = 0

Rewrite

x^{2} - 8 x + 8 y^{2} + 12 y = 0

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{( \frac{41}{2} ) ^{2}} + \frac{( y - ( - \frac{3}{4} ) ) ^{2}}{( \frac{41}{4} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, - \frac{3}{4}), a = \frac{41}{2}, b = \frac{41}{4}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (4, - \frac{3}{4}), a = \frac{41}{2}, b = \frac{41}{4}

And the center is:

(4, - \frac{3}{4})

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

9 x^{2} + 25 y^{2} = 225

x^{2} + y^{2} + 2 x - 2 x + 6 y = 0

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

\frac{( x + 3 ) ^{2}}{25} + \frac{( y - 8 ) ^{2}}{16} = 1