What is the foci ((x-1)^2)/5+((y+1)^2)/9 =1 ?

The foci ((x-1)^2)/5+((y+1)^2)/9 =1 is (1,1),(1,-3)

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foci ((x-1)^2)/5+((y+1)^2)/9 =1

foci

\frac{( x - 1 ) ^{2}}{5} + \frac{( y + 1 ) ^{2}}{9} = 1

(1, 1), (1, - 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{5} + \frac{( y + 1 ) ^{2}}{9} = 1 :

Ellipse with center

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

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

Compute

c :

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

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

Refine

(1, 1), (1, - 3)

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

\frac{( x - 2 ) ^{2}}{64} + \frac{( y - 4 ) ^{2}}{16} = 1

49 x^{2} + 25 y^{2} - 1225 = 0

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

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

What is the foci ((x-1)^2)/5+((y+1)^2)/9 =1 ?
The foci ((x-1)^2)/5+((y+1)^2)/9 =1 is (1,1),(1,-3)