What is the foci ((x-4)^2)/9+((y+6)^2)/(121)=1 ?

The foci ((x-4)^2)/9+((y+6)^2)/(121)=1 is (4,-6+4sqrt(7)),(4,-6-4sqrt(7))

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foci ((x-4)^2)/9+((y+6)^2)/(121)=1

foci

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{121} = 1

(4, - 6 + 47), (4, - 6 - 47)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{121} = 1 :

Ellipse with center

(h, k) = (4, - 6), b = 11, a = 3

(4, - 6 + c), (4, - 6 - c)

Compute

c :

c = 1 1^{2} - 3^{2} : 47

(4, - 6 + 47), (4, - 6 - 47)

\frac{( x + 7 ) ^{2}}{8} + \frac{( y - 10 ) ^{2}}{13} = 1

\frac{( x - 1 ) ^{2}}{49} + \frac{( y - 3 ) ^{2}}{36} = 1

4 x^{2} + y^{2} - 32 x + 16 y + 124 = 0

1 x^{2} + 3 y^{2} + 10 x + 8 y - 10 = 0

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

What is the foci ((x-4)^2)/9+((y+6)^2)/(121)=1 ?
The foci ((x-4)^2)/9+((y+6)^2)/(121)=1 is (4,-6+4sqrt(7)),(4,-6-4sqrt(7))