What is the foci x^2+2y^2+2x-4y-1=0 ?

The foci x^2+2y^2+2x-4y-1=0 is (-1+sqrt(2),1),(-1-sqrt(2),1)

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foci x^2+2y^2+2x-4y-1=0

foci

x^{2} + 2 y^{2} + 2 x - 4 y - 1 = 0

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

Solution steps

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

Calculate ellipse properties

x^{2} + 2 y^{2} + 2 x - 4 y - 1 = 0 :

Ellipse with center

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

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

Compute

c :

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

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

2 x^{2} + 3 y^{2} = 6

\frac{x ^{2}}{64} + \frac{y ^{2}}{1} = 1

\frac{x ^{2}}{16} + \frac{( y - 1 ) ^{2}}{36} = 1

\frac{( x + 2 ) ^{2}}{10} + \frac{( y - 5 ) ^{2}}{2} = 1

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{81} = 1

What is the foci x^2+2y^2+2x-4y-1=0 ?
The foci x^2+2y^2+2x-4y-1=0 is (-1+sqrt(2),1),(-1-sqrt(2),1)