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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

4 x^{2} + 25 y^{2} - 50 y = 75

x^{2} + 4 y^{2} - 16 = 0

16 (x - 1)^{2} + 9 (y + 2)^{2} = 144

\frac{( x - 5 ) ^{2}}{3} + \frac{( y - 7 ) ^{2}}{6} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{50} = 1

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