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

The foci ((x-2)^2}{25}+\frac{(y+1)^2)/9 =1 is (6,-1),(-2,-1)

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

foci

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

(6, - 1), (- 2, - 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, - 1), a = 5, b = 3

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

Compute

c :

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

(2 + 4, - 1), (2 - 4, - 1)

Refine

(6, - 1), (- 2, - 1)

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

5 x^{2} + 9 y^{2} + 30 x = 0

\frac{x ^{2}}{64} + \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{20} + \frac{y ^{2}}{36} = 1

\frac{x ^{2}}{12} + \frac{y ^{2}}{6} = 1

What is the foci ((x-2)^2}{25}+\frac{(y+1)^2)/9 =1 ?
The foci ((x-2)^2}{25}+\frac{(y+1)^2)/9 =1 is (6,-1),(-2,-1)