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

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

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

foci

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

(1, 3), (1, 3)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

c = 2^{2} - 2^{2} : 0

(1, 3 + 0), (1, 3 - 0)

Refine

(1, 3), (1, 3)

\frac{x ^{2}}{289} + \frac{y ^{2}}{225} = 1

4 x^{2} + y^{2} - 32 x + 4 y + 64 = 0

\frac{( x ) ^{2}}{144} + \frac{( y ) ^{2}}{169} = 1

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 1 ) ^{2}}{9} = 1

2 x^{2} + 3 y^{2} + 4 x = 1

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