What is the foci 2x^2+3y^2+4x-1=0 ?

The foci 2x^2+3y^2+4x-1=0 is (-1+sqrt(1/2),0),(-1-sqrt(1/2),0)

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foci 2x^2+3y^2+4x-1=0

foci

2 x^{2} + 3 y^{2} + 4 x - 1 = 0

(- 1 + \frac{1}{2}, 0), (- 1 - \frac{1}{2}, 0)

Solution steps

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

Calculate ellipse properties

2 x^{2} + 3 y^{2} + 4 x - 1 = 0 :

Ellipse with center

(h, k) = (- 1, 0), a = \frac{3}{2}, b = 1

(- 1 + c, 0), (- 1 - c, 0)

Compute

c :

c = \frac{3}{2}^{2} - 1^{2} : \frac{1}{2}

(- 1 + \frac{1}{2}, 0), (- 1 - \frac{1}{2}, 0)

x^{2} + \frac{( y - 6 ) ^{2}}{16} = 1

\frac{( x - 1 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

\frac{( x + 5 ) ^{2}}{20} + \frac{( y - 6 ) ^{2}}{11} = 1

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 3

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

What is the foci 2x^2+3y^2+4x-1=0 ?
The foci 2x^2+3y^2+4x-1=0 is (-1+sqrt(1/2),0),(-1-sqrt(1/2),0)