What is the foci (x^2)/4+(y^2)/(25)=1 ?

The foci (x^2)/4+(y^2)/(25)=1 is (0,sqrt(21)),(0,-sqrt(21))

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foci (x^2)/4+(y^2)/(25)=1

foci

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1

(0, 21), (0, - 21)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 2

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

Compute

c :

c = 5^{2} - 2^{2} : 21

(0, 0 + 21), (0, 0 - 21)

Refine

(0, 21), (0, - 21)

2 x - y^{2} = 0

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

x^{2} + (y - 2)^{2} = 1

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

f (x) = x^{2} - 2

What is the foci (x^2)/4+(y^2)/(25)=1 ?
The foci (x^2)/4+(y^2)/(25)=1 is (0,sqrt(21)),(0,-sqrt(21))