What is the foci ((x-3)^2)/2+((y-5)^2)/9 =1 ?

The foci ((x-3)^2)/2+((y-5)^2)/9 =1 is (3,5+sqrt(7)),(3,5-sqrt(7))

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foci ((x-3)^2)/2+((y-5)^2)/9 =1

foci

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

(3, 5 + 7), (3, 5 - 7)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (3, 5), b = 3, a = 2

(3, 5 + c), (3, 5 - c)

Compute

c :

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

(3, 5 + 7), (3, 5 - 7)

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

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

What is the foci ((x-3)^2)/2+((y-5)^2)/9 =1 ?
The foci ((x-3)^2)/2+((y-5)^2)/9 =1 is (3,5+sqrt(7)),(3,5-sqrt(7))