What is the foci (x^2)/(49)+(y^2)/(169)=1 ?

The foci (x^2)/(49)+(y^2)/(169)=1 is (0,2sqrt(30)),(0,-2sqrt(30))

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foci (x^2)/(49)+(y^2)/(169)=1

foci

\frac{x ^{2}}{49} + \frac{y ^{2}}{169} = 1

(0, 230), (0, - 230)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{49} + \frac{y ^{2}}{169} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 13, a = 7

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

Compute

c :

c = 1 3^{2} - 7^{2} : 230

(0, 0 + 230), (0, 0 - 230)

Refine

(0, 230), (0, - 230)

\frac{( y + 3 ) ^{2}}{144} + \frac{( x + 2 ) ^{2}}{25} = 1

5 x^{2} + 9 y^{2} - 20 x + 54 y + 56 = 0

25 x^{2} + 4 y^{2} - 250 x - 16 y + 541 = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{21} = 1

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

What is the foci (x^2)/(49)+(y^2)/(169)=1 ?
The foci (x^2)/(49)+(y^2)/(169)=1 is (0,2sqrt(30)),(0,-2sqrt(30))