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

The foci x^2+(y^2)/9 =1 is (0,2sqrt(2)),(0,-2sqrt(2))

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

foci

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

(0, 22), (0, - 22)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

(0, 22), (0, - 22)

\frac{( x - 2 ) ^{2}}{64} + \frac{( y - 4 ) ^{2}}{16} = 1

49 x^{2} + 25 y^{2} - 1225 = 0

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

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{( X - 1 ) ^{2}}{4} + \frac{( Y + 2 ) ^{2}}{16} = 1

What is the foci x^2+(y^2)/9 =1 ?
The foci x^2+(y^2)/9 =1 is (0,2sqrt(2)),(0,-2sqrt(2))