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

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

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foci (x^2}{25}+\frac{y^2)/4 =1

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

y = - \frac{x ^{2}}{10} + \frac{9 x}{10} + \frac{11}{5}

2 y^{2} - 12 y - x + 5 = 0

9 x^{2} + 9 y^{2} + 18 x - 18 y + 14 = 0

x^{2} = 8 y

y^{2} + 4 x = 7

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