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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

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

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{11} = 1

\frac{x ^{2}}{256} + \frac{y ^{2}}{400} = 1

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

What is the foci x^2+2y^2-2x+8y+5=0 ?
The foci x^2+2y^2-2x+8y+5=0 is (1+sqrt(2),-2),(1-sqrt(2),-2)