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

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

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

foci

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

(- 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 :

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)

\frac{x ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

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

\frac{x ^{2}}{576} + \frac{y ^{2}}{676} = 1

x^{2} + 3 y^{2} + 6 x + 6 = 0

16 x^{2} + 25 y^{2} - 48 x + 100 y - 264 = 0

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