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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(1 + c, - 2), (1 - c, - 2)

Compute

c :

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

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

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

16 x^{2} + 9 y^{2} - 144 = 0

x^{2} + 3 y^{2} = 9

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

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