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

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

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

foci

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

(- 2 + 21, 4), (- 2 - 21, 4)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(- 2 + c, 4), (- 2 - c, 4)

Compute

c :

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

(- 2 + 21, 4), (- 2 - 21, 4)

2 x^{2} + y^{2} = 50

\frac{( x - 5 ) ^{2}}{8 ^{2}} + \frac{( y + 2 ) ^{2}}{1 0 ^{2}} = 1

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

9 x^{2} - 54 x + 4 y^{2} = - 45

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{144} = 1

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