What is the foci ((x-5)^2)/3+((y-7)^2)/6 =1 ?

The foci ((x-5)^2)/3+((y-7)^2)/6 =1 is (5,7+sqrt(3)),(5,7-sqrt(3))

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foci ((x-5)^2)/3+((y-7)^2)/6 =1

foci

\frac{( x - 5 ) ^{2}}{3} + \frac{( y - 7 ) ^{2}}{6} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{3} + \frac{( y - 7 ) ^{2}}{6} = 1 :

Ellipse with center

(h, k) = (5, 7), b = 6, a = 3

(5, 7 + c), (5, 7 - c)

Compute

c :

c = 6^{2} - 3^{2} : 3

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

\frac{x ^{2}}{100} + \frac{y ^{2}}{50} = 1

\frac{x ^{2}}{49} + \frac{y ^{2}}{81} = 1

25 x^{2} + 16 y^{2} + 100 x + 96 y - 156 = 0

9 x^{2} + 16 y^{2} + 64 y - 80 = 0

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

What is the foci ((x-5)^2)/3+((y-7)^2)/6 =1 ?
The foci ((x-5)^2)/3+((y-7)^2)/6 =1 is (5,7+sqrt(3)),(5,7-sqrt(3))