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

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

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

foci

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

(- 1, 5), (- 1, - 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 1, 2), b = 5, a = 4

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

Compute

c :

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

(- 1, 2 + 3), (- 1, 2 - 3)

Refine

(- 1, 5), (- 1, - 1)

9 (x - 5)^{2} + 4 (y + 2)^{2} = 36

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{3} = 1

y^{2} = 1 - 25 x^{2}

2 x^{2} - 8 x + 3 y^{2} + 6 y + 5 = 0

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