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

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

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

foci

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

(- 5 + 22, 1), (- 5 - 22, 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(- 5 + c, 1), (- 5 - c, 1)

Compute

c :

c = 3^{2} - 1^{2} : 22

(- 5 + 22, 1), (- 5 - 22, 1)

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

\frac{x ^{2}}{49} + \frac{y ^{2}}{9} = 1

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

7 x^{2} + 16 y^{2} = 112

\frac{x ^{2}}{5} + \frac{y ^{2}}{3} = 1

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