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

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

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

foci

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

(0, 2), (- 2, 2)

Solution steps

(h + c, k), (h - c, k)

Calculate ellipse properties

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

Ellipse with center

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

(- 1 + c, 2), (- 1 - c, 2)

Compute

c :

c = 2^{2} - 1^{2} : 1

(- 1 + 1, 2), (- 1 - 1, 2)

Refine

(0, 2), (- 2, 2)

\frac{( x + 13 ) ^{2}}{400} + \frac{( y + 6 ) ^{2}}{625} = 1

4 x^{2} + y^{2} - 16 x - 2 y - 19 = 0

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

3 x^{2} + 7 y^{2} = 63

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

What is the foci x^2+2y^2+2x-8y+7=0 ?
The foci x^2+2y^2+2x-8y+7=0 is (0,2),(-2,2)