What is the foci (x^2}{25}+\frac{y^2)/9 =1 ?

The foci (x^2}{25}+\frac{y^2)/9 =1 is (4,0),(-4,0)

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foci (x^2}{25}+\frac{y^2)/9 =1

foci

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1

(4, 0), (- 4, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 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} : 4

(0 + 4, 0), (0 - 4, 0)

Refine

(4, 0), (- 4, 0)

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

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

x^{2} - y^{2} = 3

9 x^{2} + 4 y^{2} - 18 x + 16 y - 11 = 0

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

What is the foci (x^2}{25}+\frac{y^2)/9 =1 ?
The foci (x^2}{25}+\frac{y^2)/9 =1 is (4,0),(-4,0)