What is the foci ((x)^2)/(44)+(y^2)/(125)=1 ?

The foci ((x)^2)/(44)+(y^2)/(125)=1 is (0,9),(0,-9)

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foci ((x)^2)/(44)+(y^2)/(125)=1

foci

\frac{( x ) ^{2}}{44} + \frac{y ^{2}}{125} = 1

(0, 9), (0, - 9)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{44} + \frac{y ^{2}}{125} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 55, a = 211

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

Compute

c :

c = (55)^{2} - (211)^{2} : 9

(0, 0 + 9), (0, 0 - 9)

Refine

(0, 9), (0, - 9)

2 x^{2} + 8 y^{2} = 16

64 x^{2} + y^{2} = 64

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

25 x^{2} + 16 y^{2} + 100 x - 300 = 0

\frac{x ^{2}}{121} + \frac{( y + 4 ) ^{2}}{23} = 1

What is the foci ((x)^2)/(44)+(y^2)/(125)=1 ?
The foci ((x)^2)/(44)+(y^2)/(125)=1 is (0,9),(0,-9)