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

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

(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)

(\frac{x}{2.5})^{2} + (\frac{y}{3})^{2} = 1

9 x^{2} + 4 y^{2} + 54 x - 8 y + 49 = 0

25 (x - 1)^{2} + 9 (y + 4)^{2} = 225

\frac{( x + 1 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

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)