What is the foci (x^2)/(36)+((y+1)^2)/(25)=1 ?

The foci (x^2)/(36)+((y+1)^2)/(25)=1 is (sqrt(11),-1),(-sqrt(11),-1)

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foci (x^2)/(36)+((y+1)^2)/(25)=1

foci

\frac{x ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

(11, - 1), (- 11, - 1)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, - 1), a = 6, b = 5

(0 + c, - 1), (0 - c, - 1)

Compute

c :

c = 6^{2} - 5^{2} : 11

(0 + 11, - 1), (0 - 11, - 1)

Refine

(11, - 1), (- 11, - 1)

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

\frac{x ^{2}}{576} + \frac{y ^{2}}{676} = 1

x^{2} + 3 y^{2} + 6 x + 6 = 0

16 x^{2} + 25 y^{2} - 48 x + 100 y - 264 = 0

9 x^{2} + 25 y^{2} + 54 x + 100 y - 44 = 0

What is the foci (x^2)/(36)+((y+1)^2)/(25)=1 ?
The foci (x^2)/(36)+((y+1)^2)/(25)=1 is (sqrt(11),-1),(-sqrt(11),-1)