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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

9 x^{2} + 16 y^{2} + 72 x + 128 y + 256 = 0

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1

\frac{x ^{2}}{15} + \frac{y ^{2}}{64} = 1

25 x^{2} + 9 y^{2} - 150 x - 18 y + 9 = 0

\frac{x ^{2}}{64} + \frac{y ^{2}}{4} = 1

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