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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 1), a = 2, b = 1

(0 + c, 1), (0 - c, 1)

Compute

c :

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

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

Refine

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

3 x^{2} + 7 y^{2} = 63

25 x^{2} + 9 y^{2} = 225

\frac{x ^{2}}{34} + \frac{y ^{2}}{9} = 1

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

9 x^{2} + 4 y^{2} - 36 x + 40 y + 100 = 0

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