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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

36 x^{2} + 9 y^{2} + 48 x - 36 y + 43 = 0

4 x^{2} + 9 y^{2} - 18 y - 27 = 0

9 x^{2} + 36 x + y^{2} + 2 y + 36 = 0

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

2 x^{2} + 5 y^{2} = 10

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