What is the foci (x^2)/(25)+((y+3)^2)/(49)=1 ?

The foci (x^2)/(25)+((y+3)^2)/(49)=1 is (0,-3+2sqrt(6)),(0,-3-2sqrt(6))

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foci (x^2)/(25)+((y+3)^2)/(49)=1

foci

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1

(0, - 3 + 26), (0, - 3 - 26)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, - 3), b = 7, a = 5

(0, - 3 + c), (0, - 3 - c)

Compute

c :

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

(0, - 3 + 26), (0, - 3 - 26)

3 x^{2} + 2 y^{2} + 6 x - 4 y + 5 = 18

\frac{x ^{2}}{49} + \frac{y ^{2}}{25} = 1

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{169} = 1

\frac{( x - 3 ) ^{2}}{25} + \frac{y ^{2}}{9} = 1

x^{2} + 4 y^{2} = 16

What is the foci (x^2)/(25)+((y+3)^2)/(49)=1 ?
The foci (x^2)/(25)+((y+3)^2)/(49)=1 is (0,-3+2sqrt(6)),(0,-3-2sqrt(6))