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

The foci ((x+3)^2)/9+((y-1)^2)/(25)=1 is (-3,5),(-3,-3)

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foci ((x+3)^2)/9+((y-1)^2)/(25)=1

foci

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

(- 3, 5), (- 3, - 3)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

(- 3, 5), (- 3, - 3)

4 x^{2} + 9 y^{2} - 48 x + 72 y = - 144

36 x^{2} + 16 y^{2} - 180 x + 96 y - 207 = 0

x^{2} + 2 y^{2} + 2 x - 4 y - 1 = 0

2 x^{2} + 3 y^{2} = 6

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

What is the foci ((x+3)^2)/9+((y-1)^2)/(25)=1 ?
The foci ((x+3)^2)/9+((y-1)^2)/(25)=1 is (-3,5),(-3,-3)