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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

6 x^{2} + 8 y^{2} + 12 x - 32 y = 178

9 x^{2} + 49 y^{2} - 144 x + 490 y + 1360 = 0

9 x^{2} + 18 x + 4 y^{2} - 24 y + 9 = 0

x^{2} + 4 y^{2} - 6 x + 20 y - 2 = 0

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

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