What is the foci (x^2)/3+(y^2)/5 =1 ?

The foci (x^2)/3+(y^2)/5 =1 is (0,sqrt(2)),(0,-sqrt(2))

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foci (x^2)/3+(y^2)/5 =1

foci

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 3

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

Compute

c :

c = 5^{2} - 3^{2} : 2

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

Refine

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

16 x^{2} + 25 y^{2} - 64 x + 150 y - 111 = 0

9 x^{2} + 49 y^{2} = 441

\frac{( x - 1 ) ^{2}}{425} + \frac{( y - 2 ) ^{2}}{400} = 1

9 x^{2} + 4 y^{2} + 18 x - 8 y - 131 = 0

\frac{( x - 2 ) ^{2}}{225} + \frac{( y - 8 ) ^{2}}{81} = 1

What is the foci (x^2)/3+(y^2)/5 =1 ?
The foci (x^2)/3+(y^2)/5 =1 is (0,sqrt(2)),(0,-sqrt(2))