What is the foci ((x+1)^2)/(25)+((y-4)^2)/(169)=1 ?

The foci ((x+1)^2)/(25)+((y-4)^2)/(169)=1 is (-1,16),(-1,-8)

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foci ((x+1)^2)/(25)+((y-4)^2)/(169)=1

foci

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{169} = 1

(- 1, 16), (- 1, - 8)

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{169} = 1 :

Ellipse with center

(h, k) = (- 1, 4), b = 13, a = 5

(- 1, 4 + c), (- 1, 4 - c)

Compute

c :

c = 1 3^{2} - 5^{2} : 12

(- 1, 4 + 12), (- 1, 4 - 12)

Refine

(- 1, 16), (- 1, - 8)

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

\frac{x ^{2}}{121} + \frac{( y + 4 ) ^{2}}{1} = 1

9 x^{2} + 5 y^{2} = 45

16 x^{2} + 25 y^{2} + 64 x - 336 = 0

\frac{( x - 7 ) ^{2}}{81} + \frac{( y - 6 ) ^{2}}{64} = 1

What is the foci ((x+1)^2)/(25)+((y-4)^2)/(169)=1 ?
The foci ((x+1)^2)/(25)+((y-4)^2)/(169)=1 is (-1,16),(-1,-8)