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

The foci ((x-1)^2)/9+((y+2)^2)/5 =1 is (3,-2),(-1,-2)

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foci ((x-1)^2)/9+((y+2)^2)/5 =1

foci

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{5} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{5} = 1 :

Ellipse with center

(h, k) = (1, - 2), a = 3, b = 5

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

Compute

c :

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

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

Refine

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

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

4 x^{2} + 9 y^{2} + 16 x - 90 y + 205 = 0

9 x^{2} + 4 y^{2} - 18 x - 16 y - 11 = 0

\frac{( x - 1 ) ^{2}}{36} + \frac{( y - 2 ) ^{2}}{115} = 3

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

What is the foci ((x-1)^2)/9+((y+2)^2)/5 =1 ?
The foci ((x-1)^2)/9+((y+2)^2)/5 =1 is (3,-2),(-1,-2)