What is the foci (x^2)/1+(y^2)/5 =1 ?

The foci (x^2)/1+(y^2)/5 =1 is (0,2),(0,-2)

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foci (x^2)/1+(y^2)/5 =1

foci

\frac{x ^{2}}{1} + \frac{y ^{2}}{5} = 1

(0, 2), (0, - 2)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{1} + \frac{y ^{2}}{5} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 1

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

Compute

c :

c = 5^{2} - 1^{2} : 2

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

Refine

(0, 2), (0, - 2)

\frac{( y + 3 ) ^{2}}{169} + \frac{( x + 2 ) ^{2}}{25} = 1

\frac{x ^{2}}{15} + \frac{y ^{2}}{45} = 1

9 x^{2} + 16 y^{2} + 96 y = 0

9 x^{2} + 4 y^{2} - 18 x + 16 y - 11 = 0

\frac{x ^{2}}{99} + \frac{y ^{2}}{225} = 1

What is the foci (x^2)/1+(y^2)/5 =1 ?
The foci (x^2)/1+(y^2)/5 =1 is (0,2),(0,-2)