What is the foci ((x-3)^2}{25}+\frac{(y+1)^2)/9 =1 ?

The foci ((x-3)^2}{25}+\frac{(y+1)^2)/9 =1 is (7,-1),(-1,-1)

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foci ((x-3)^2}{25}+\frac{(y+1)^2)/9 =1

foci

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{9} = 1

(7, - 1), (- 1, - 1)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (3, - 1), a = 5, b = 3

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

Compute

c :

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

(3 + 4, - 1), (3 - 4, - 1)

Refine

(7, - 1), (- 1, - 1)

9 x^{2} + 16 y^{2} - 36 x + 96 y + 36 = 0

\frac{( x - 1 ) ^{2}}{45} + \frac{y ^{2}}{54} = 1

36 x^{2} + 5 y^{2} - 90 y - 495 = 0

16 x^{2} - 32 x + 9 y^{2} = 128

(\frac{x ^{2}}{15}) + (\frac{y ^{2}}{16}) = 1

What is the foci ((x-3)^2}{25}+\frac{(y+1)^2)/9 =1 ?
The foci ((x-3)^2}{25}+\frac{(y+1)^2)/9 =1 is (7,-1),(-1,-1)