What is the foci (x^2)/(16)+(y^2)/(25)=1 ?

The foci (x^2)/(16)+(y^2)/(25)=1 is (0,3),(0,-3)

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foci (x^2)/(16)+(y^2)/(25)=1

foci

\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 4

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

Compute

c :

c = 5^{2} - 4^{2} : 3

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

Refine

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

What is the foci (x^2)/(16)+(y^2)/(25)=1 ?
The foci (x^2)/(16)+(y^2)/(25)=1 is (0,3),(0,-3)