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

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

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

foci

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

(0, - 4 + 21), (0, - 4 - 21)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, - 4), b = 5, a = 2

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

Compute

c :

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

(0, - 4 + 21), (0, - 4 - 21)

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

\frac{y ^{2}}{16} - \frac{x ^{2}}{9} = 1

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

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

- 2 y^{2} + x - 4 y + 1 = 0

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