What is the center x^2+2y^2-2x+8y+5=0 ?

The center x^2+2y^2-2x+8y+5=0 is (1,-2)

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center x^2+2y^2-2x+8y+5=0

center

x^{2} + 2 y^{2} - 2 x + 8 y + 5 = 0

(1, - 2)

Solution steps

x^{2} + 2 y^{2} - 2 x + 8 y + 5 = 0

Rewrite

x^{2} + 2 y^{2} - 2 x + 8 y + 5 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(1, - 2)

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1

3 x^{2} + 2 y^{2} + 6 x - 4 y + 5 = 18

4 x^{2} + 9 y^{2} - 16 x + 18 y - 11 = 0

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{36} = 1