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

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

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

foci

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

(1, - 2), (1, - 2)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

(1, - 2 + 0), (1, - 2 - 0)

Refine

(1, - 2), (1, - 2)

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{25} = 1

\frac{x ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{x ^{2}}{9} + \frac{y ^{2}}{81} = 1

16 xx + 25 yy = 400

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

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