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

The foci ((x+1)^2)/(25)+(y^2)/(16)=1 is (2,0),(-4,0)

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

foci

(2, 0), (- 4, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 1, 0), a = 5, b = 4

(- 1 + c, 0), (- 1 - c, 0)

Compute

c :

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

(- 1 + 3, 0), (- 1 - 3, 0)

Refine

(2, 0), (- 4, 0)

What is the foci ((x+1)^2)/(25)+(y^2)/(16)=1 ?
The foci ((x+1)^2)/(25)+(y^2)/(16)=1 is (2,0),(-4,0)