What is the foci 9(x+2)^2+(y+1)^2=1 ?

The foci 9(x+2)^2+(y+1)^2=1 is (-2,(-3+2sqrt(2))/3),(-2,(-3-2sqrt(2))/3)

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foci 9(x+2)^2+(y+1)^2=1

foci

9 (x + 2)^{2} + (y + 1)^{2} = 1

(- 2, \frac{- 3 + 2 2}{3}), (- 2, \frac{- 3 - 2 2}{3})

Solution steps

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

Calculate ellipse properties

9 (x + 2)^{2} + (y + 1)^{2} = 1 :

Ellipse with center

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

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

Compute

c :

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

(- 2, - 1 + \frac{2 2}{3}), (- 2, - 1 - \frac{2 2}{3})

Refine

(- 2, \frac{- 3 + 2 2}{3}), (- 2, \frac{- 3 - 2 2}{3})

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

9 x^{2} + 25 y^{2} + 90 x = 0

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{9 ^{2}} = 1

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{16} = 1

27 x^{2} + 48 y^{2} - 54 x - 192 y - 213 = 0

What is the foci 9(x+2)^2+(y+1)^2=1 ?
The foci 9(x+2)^2+(y+1)^2=1 is (-2,(-3+2sqrt(2))/3),(-2,(-3-2sqrt(2))/3)