What is the foci ((x+3)^2)/(16)+((y-5)^2)/(25)=1 ?

The foci ((x+3)^2)/(16)+((y-5)^2)/(25)=1 is (-3,8),(-3,2)

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foci ((x+3)^2)/(16)+((y-5)^2)/(25)=1

foci

\frac{( x + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

(- 3, 8), (- 3, 2)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (- 3, 5), b = 5, a = 4

(- 3, 5 + c), (- 3, 5 - c)

Compute

c :

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

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

Refine

(- 3, 8), (- 3, 2)

x^{2} - 8 x + y^{2} + 2 y = - 16

x^{2} = 12 y

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

16 x^{2} + 9 y^{2} = 144

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

What is the foci ((x+3)^2)/(16)+((y-5)^2)/(25)=1 ?
The foci ((x+3)^2)/(16)+((y-5)^2)/(25)=1 is (-3,8),(-3,2)