What is the foci ((x-4)^2)/9+((y+2)^2)/(16)=1 ?

The foci ((x-4)^2)/9+((y+2)^2)/(16)=1 is (4,-2+sqrt(7)),(4,-2-sqrt(7))

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foci ((x-4)^2)/9+((y+2)^2)/(16)=1

foci

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{16} = 1

(4, - 2 + 7), (4, - 2 - 7)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (4, - 2), b = 4, a = 3

(4, - 2 + c), (4, - 2 - c)

Compute

c :

c = 4^{2} - 3^{2} : 7

(4, - 2 + 7), (4, - 2 - 7)

27 x^{2} + 48 y^{2} - 54 x - 192 y - 213 = 0

\frac{x ^{2}}{46} + \frac{( y + 8 ) ^{2}}{26} = 1

2 x^{2} + 3 y^{2} + 4 x - 6 y - 1 = 0

\frac{x ^{2}}{15} + \frac{( y - 5 ) ^{2}}{20} = 1

x^{2} + 9 y^{2} - 8 x + 18 y - 11 = 0

What is the foci ((x-4)^2)/9+((y+2)^2)/(16)=1 ?
The foci ((x-4)^2)/9+((y+2)^2)/(16)=1 is (4,-2+sqrt(7)),(4,-2-sqrt(7))