What is the foci (x^2)/9+(y^2)/3 =1 ?

The foci (x^2)/9+(y^2)/3 =1 is (sqrt(6),0),(-sqrt(6),0)

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foci (x^2)/9+(y^2)/3 =1

foci

\frac{x ^{2}}{9} + \frac{y ^{2}}{3} = 1

(6, 0), (- 6, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{9} + \frac{y ^{2}}{3} = 1 :

Ellipse with center

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

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

Compute

c :

c = 3^{2} - 3^{2} : 6

(0 + 6, 0), (0 - 6, 0)

Refine

(6, 0), (- 6, 0)

y^{2} = 1 - 25 x^{2}

2 x^{2} - 8 x + 3 y^{2} + 6 y + 5 = 0

36 x^{2} + 288 x + 64 y^{2} - 256 y = 1472

\frac{x ^{2}}{81} + \frac{y ^{2}}{56} = 1

x^{2} - 4 x + 9 y^{2} + 54 y + 4 = 0

What is the foci (x^2)/9+(y^2)/3 =1 ?
The foci (x^2)/9+(y^2)/3 =1 is (sqrt(6),0),(-sqrt(6),0)