What is the center x^2+2y^2+2x-8y+7=0 ?

The center x^2+2y^2+2x-8y+7=0 is (-1,2)

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center x^2+2y^2+2x-8y+7=0

center

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

(- 1, 2)

Solution steps

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

Rewrite

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(- 1, 2)

4 x^{2} + 25 y^{2} = 100

2 (x + \frac{1}{4})^{2} + y^{2} = 8

x^{2} + 4 y^{2} = 16

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

4 x^{2} + y^{2} = 1