What is the foci ((x-4)^2)/3+(y^2)/9 =1 ?

The foci ((x-4)^2)/3+(y^2)/9 =1 is (4,sqrt(6)),(4,-sqrt(6))

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foci ((x-4)^2)/3+(y^2)/9 =1

foci

\frac{( x - 4 ) ^{2}}{3} + \frac{y ^{2}}{9} = 1

(4, 6), (4, - 6)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{3} + \frac{y ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (4, 0), b = 3, a = 3

(4, 0 + c), (4, 0 - c)

Compute

c :

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

(4, 0 + 6), (4, 0 - 6)

Refine

(4, 6), (4, - 6)

\frac{x ^{2}}{36} + \frac{8 y ^{2}}{81} = 1

x^{2} + 3 y^{2} + 8 x + 9 y - 14 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{13} = 1

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

25 \frac{x ^{2}}{4} - 50 x + 100 y^{2} + 800 y + 1075 = 0

What is the foci ((x-4)^2)/3+(y^2)/9 =1 ?
The foci ((x-4)^2)/3+(y^2)/9 =1 is (4,sqrt(6)),(4,-sqrt(6))