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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

c = 2^{2} - 1^{2} : 3

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

37 x^{2} + y^{2} = 37

\frac{( x - 1 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{52} = 1

\frac{x ^{2}}{5} + \frac{y ^{2}}{4} = 1

\frac{( x + 20 ) ^{2}}{105} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{12} + \frac{( y ) ^{2}}{16} = 1

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