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

The foci ((x+2)^2)/(25)+((y-1)^2)/(16)=1 is (1,1),(-5,1)

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

foci

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

(1, 1), (- 5, 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, 1), a = 5, b = 4

(- 2 + c, 1), (- 2 - c, 1)

Compute

c :

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

(- 2 + 3, 1), (- 2 - 3, 1)

Refine

(1, 1), (- 5, 1)

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

16 x^{2} + 9 y^{2} - 144 = 0

x^{2} + 3 y^{2} = 9

x^{2} + 3 y^{2} = 3

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1

What is the foci ((x+2)^2)/(25)+((y-1)^2)/(16)=1 ?
The foci ((x+2)^2)/(25)+((y-1)^2)/(16)=1 is (1,1),(-5,1)