What is the foci ((x-4)^2)/9+(y+3)^2=1 ?

The foci ((x-4)^2)/9+(y+3)^2=1 is (4+2sqrt(2),-3),(4-2sqrt(2),-3)

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foci ((x-4)^2)/9+(y+3)^2=1

foci

\frac{( x - 4 ) ^{2}}{9} + (y + 3)^{2} = 1

(4 + 22, - 3), (4 - 22, - 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{9} + (y + 3)^{2} = 1 :

Ellipse with center

(h, k) = (4, - 3), a = 3, b = 1

(4 + c, - 3), (4 - c, - 3)

Compute

c :

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

(4 + 22, - 3), (4 - 22, - 3)

2 x^{2} + 5 y^{2} = 10

x^{2} + 4 y^{2} - 2 x - 16 y + 13 = 0

4 x^{2} + 9 y^{2} = 25

4 y^{2} + 2 x^{2} - 4 y - 8 x - 15 = 0

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

What is the foci ((x-4)^2)/9+(y+3)^2=1 ?
The foci ((x-4)^2)/9+(y+3)^2=1 is (4+2sqrt(2),-3),(4-2sqrt(2),-3)