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

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

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

foci

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

(2 + 11, - 1), (2 - 11, - 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, - 1), a = 6, b = 5

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

Compute

c :

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

(2 + 11, - 1), (2 - 11, - 1)

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1

9 x^{2} - y^{2} - 36 x - 6 y + 18 = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{49} = 1

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

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