What is the foci (x^2)/8+(y^2)/5 =1 ?

The foci (x^2)/8+(y^2)/5 =1 is (sqrt(3),0),(-sqrt(3),0)

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foci (x^2)/8+(y^2)/5 =1

foci

\frac{x ^{2}}{8} + \frac{y ^{2}}{5} = 1

(3, 0), (- 3, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{8} + \frac{y ^{2}}{5} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 22, b = 5

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

Compute

c :

c = (22)^{2} - 5^{2} : 3

(0 + 3, 0), (0 - 3, 0)

Refine

(3, 0), (- 3, 0)

16 x^{2} + 25 y^{2} - 96 x + 100 y - 156 = 0

+ \frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{25} = 1

81 x^{2} + 64 y^{2} = 5184

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

What is the foci (x^2)/8+(y^2)/5 =1 ?
The foci (x^2)/8+(y^2)/5 =1 is (sqrt(3),0),(-sqrt(3),0)