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

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

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foci 1/4 (x-5)^2+1/9 (y+5)^2=1

foci

\frac{1}{4} (x - 5)^{2} + \frac{1}{9} (y + 5)^{2} = 1

(5, - 5 + 5), (5, - 5 - 5)

Solution steps

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

Calculate ellipse properties

\frac{1}{4} (x - 5)^{2} + \frac{1}{9} (y + 5)^{2} = 1 :

Ellipse with center

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

(5, - 5 + c), (5, - 5 - c)

Compute

c :

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

(5, - 5 + 5), (5, - 5 - 5)

\frac{( x + 1 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{49} = 1

4 x^{2} + 36 y^{2} = 144

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

1 x^{2} + 4 y^{2} + 10 x + 9 y - 10 = 0

\frac{( x + 2 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{25} = 1

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