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

The foci (2x^2)/(25)+((y-3)^2)/(49)=1 is (0,3+sqrt(73/2)),(0,3-sqrt(73/2))

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

foci

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

(0, 3 + \frac{73}{2}), (0, 3 - \frac{73}{2})

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 3), b = 7, a = \frac{5}{2}

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

Compute

c :

c = 7^{2} - (\frac{5}{2})^{2} : \frac{73}{2}

(0, 3 + \frac{73}{2}), (0, 3 - \frac{73}{2})

\frac{( x - 3 ) ^{2}}{2} + \frac{( y - 5 ) ^{2}}{9} = 1

4 x^{2} + 81 y^{2} = 324

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{4} = 1

7 x^{2} = 21 - 3 y^{2}

9 x^{2} + 25 y^{2} + 18 x + 50 y - 191 = 0

What is the foci (2x^2)/(25)+((y-3)^2)/(49)=1 ?
The foci (2x^2)/(25)+((y-3)^2)/(49)=1 is (0,3+sqrt(73/2)),(0,3-sqrt(73/2))