What is the foci (x^2)/9+(y^2)/5 =1 ?

The foci (x^2)/9+(y^2)/5 =1 is (2,0),(-2,0)

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foci (x^2)/9+(y^2)/5 =1

foci

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1 :

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

y^{2} = 5 x

9 x^{2} - 4 y^{2} = 36

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

y^{2} - 4 y + 4 x = - 16

x^{2} + y^{2} + 6 x - 4 y - 23 = 0

What is the foci (x^2)/9+(y^2)/5 =1 ?
The foci (x^2)/9+(y^2)/5 =1 is (2,0),(-2,0)