What is the foci 49x^2+25y^2=1225 ?

The foci 49x^2+25y^2=1225 is (0,2sqrt(6)),(0,-2sqrt(6))

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foci 49x^2+25y^2=1225

foci

49 x^{2} + 25 y^{2} = 1225

(0, 26), (0, - 26)

Solution steps

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

Calculate ellipse properties

49 x^{2} + 25 y^{2} = 1225 :

Ellipse with center

(h, k) = (0, 0), b = 7, a = 5

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

Compute

c :

c = 7^{2} - 5^{2} : 26

(0, 0 + 26), (0, 0 - 26)

Refine

(0, 26), (0, - 26)

\frac{x ^{2}}{64} + \frac{y ^{2}}{49} = 1

3 x^{2} + 12 y^{2} - 6 x - 48 y + 33 = 0

16 x^{2} - 64 x + 9 y^{2} = 80

4 x^{2} + 9 y^{2} - 8 x - 32 = 0

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1

What is the foci 49x^2+25y^2=1225 ?
The foci 49x^2+25y^2=1225 is (0,2sqrt(6)),(0,-2sqrt(6))