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

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

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

foci

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

(3, 3), (- 1, 3)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (1, 3), a = 6, b = 2

(1 + c, 3), (1 - c, 3)

Compute

c :

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

(1 + 2, 3), (1 - 2, 3)

Refine

(3, 3), (- 1, 3)

\frac{x ^{2}}{81} + \frac{y ^{2}}{225} = 1

\frac{( x + 4 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{16} = 1

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{36} = 1

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

\frac{x ^{2}}{5} + \frac{y ^{2}}{16} = 1

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