What is the foci ((x+1)^2)/(16)+((y+1)^2)/(25)=1 ?

The foci ((x+1)^2)/(16)+((y+1)^2)/(25)=1 is (-1,2),(-1,-4)

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foci ((x+1)^2)/(16)+((y+1)^2)/(25)=1

foci

\frac{( x + 1 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{25} = 1

(- 1, 2), (- 1, - 4)

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (- 1, - 1), b = 5, a = 4

(- 1, - 1 + c), (- 1, - 1 - c)

Compute

c :

c = 5^{2} - 4^{2} : 3

(- 1, - 1 + 3), (- 1, - 1 - 3)

Refine

(- 1, 2), (- 1, - 4)

\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1

4 x^{2} + y^{2} - 8 x + 4 y - 8 = 0

x^{2} + 2 y^{2} - 2 x + 8 y + 5 = 0

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

3 x^{2} + 2 y^{2} + 6 x - 4 y + 5 = 18

What is the foci ((x+1)^2)/(16)+((y+1)^2)/(25)=1 ?
The foci ((x+1)^2)/(16)+((y+1)^2)/(25)=1 is (-1,2),(-1,-4)