What is the foci ((x-2)^2)/9+((y-1)^2)/4 =1 ?

The foci ((x-2)^2)/9+((y-1)^2)/4 =1 is (2+sqrt(5),1),(2-sqrt(5),1)

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foci ((x-2)^2)/9+((y-1)^2)/4 =1

foci

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

(2 + 5, 1), (2 - 5, 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

c = 3^{2} - 2^{2} : 5

(2 + 5, 1), (2 - 5, 1)

\frac{( x - 4 ) ^{2}}{24} + \frac{( y - 2 ) ^{2}}{49} = 1

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

\frac{( x + 13 ) ^{2}}{400} + \frac{( y + 6 ) ^{2}}{625} = 1

4 x^{2} + y^{2} - 16 x - 2 y - 19 = 0

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

What is the foci ((x-2)^2)/9+((y-1)^2)/4 =1 ?
The foci ((x-2)^2)/9+((y-1)^2)/4 =1 is (2+sqrt(5),1),(2-sqrt(5),1)