What is the foci ((x+5)^2)/4+((y+5)^2)/(25)=1 ?

The foci ((x+5)^2)/4+((y+5)^2)/(25)=1 is (-5,-5+sqrt(21)),(-5,-5-sqrt(21))

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foci ((x+5)^2)/4+((y+5)^2)/(25)=1

foci

\frac{( x + 5 ) ^{2}}{4} + \frac{( y + 5 ) ^{2}}{25} = 1

(- 5, - 5 + 21), (- 5, - 5 - 21)

Solution steps

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

Calculate ellipse properties

\frac{( x + 5 ) ^{2}}{4} + \frac{( y + 5 ) ^{2}}{25} = 1 :

Ellipse with center

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

(- 5, - 5 + c), (- 5, - 5 - c)

Compute

c :

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

(- 5, - 5 + 21), (- 5, - 5 - 21)

4 x^{2} + 9 y^{2} + 16 x + 18 y - 11 = 0

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1

4 x^{2} + 25 y^{2} + 40 x - 150 y + 225 = 0

\frac{x ^{2}}{\frac{81}{4}} + \frac{y ^{2}}{\frac{25}{16}} = 1

2 x^{2} + 3 y^{2} + 4 x - 1 = 0

What is the foci ((x+5)^2)/4+((y+5)^2)/(25)=1 ?
The foci ((x+5)^2)/4+((y+5)^2)/(25)=1 is (-5,-5+sqrt(21)),(-5,-5-sqrt(21))