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

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

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

foci

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

(22, 1), (- 22, 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

(0 + 22, 1), (0 - 22, 1)

Refine

(22, 1), (- 22, 1)

\frac{x ^{2}}{8} + \frac{y ^{2}}{24} = 1

\frac{( x - 2 ) ^{2}}{49} + \frac{( y + 1 ) ^{2}}{25} = 1

25 x^{2} + 9 y^{2} + 100 x - 125 = 0

4 x^{2} + 25 y^{2} - 40 x + 150 y + 225 = 0

\frac{( x - 8 ) ^{2}}{64} + \frac{( y - 5 ) ^{2}}{5 ^{2}} = 1

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