What is the foci (x^2)/(64)+(y^2)/(121)=1 ?

The foci (x^2)/(64)+(y^2)/(121)=1 is (0,sqrt(57)),(0,-sqrt(57))

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foci (x^2)/(64)+(y^2)/(121)=1

foci

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1

(0, 57), (0, - 57)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 11, a = 8

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

Compute

c :

c = 1 1^{2} - 8^{2} : 57

(0, 0 + 57), (0, 0 - 57)

Refine

(0, 57), (0, - 57)

9 x^{2} + 25 y^{2} + 18 x - 50 y = 191

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

4 x^{2} + 16 y^{2} - 4 x - 32 y + 1 = 0

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{5100} = 1

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1

What is the foci (x^2)/(64)+(y^2)/(121)=1 ?
The foci (x^2)/(64)+(y^2)/(121)=1 is (0,sqrt(57)),(0,-sqrt(57))