What is the foci x^2+9y^2+4x=5 ?

The foci x^2+9y^2+4x=5 is (-2+2sqrt(2),0),(-2-2sqrt(2),0)

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foci x^2+9y^2+4x=5

foci

x^{2} + 9 y^{2} + 4 x = 5

(- 2 + 22, 0), (- 2 - 22, 0)

Solution steps

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

Calculate ellipse properties

x^{2} + 9 y^{2} + 4 x = 5 :

Ellipse with center

(h, k) = (- 2, 0), a = 3, b = 1

(- 2 + c, 0), (- 2 - c, 0)

Compute

c :

c = 3^{2} - 1^{2} : 22

(- 2 + 22, 0), (- 2 - 22, 0)

4 x^{2} - 40 x + 2 y^{2} - 20 y = - 142

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

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 1 ) ^{2}}{49} = 1

\frac{( x + 20 ) ^{2}}{16} + \frac{y ^{2}}{89} = 1

\frac{x ^{2}}{\frac{9}{4}} + \frac{y ^{2}}{\frac{25}{4}} = 1

What is the foci x^2+9y^2+4x=5 ?
The foci x^2+9y^2+4x=5 is (-2+2sqrt(2),0),(-2-2sqrt(2),0)