What is the foci ((x+2)^2)/(16)+((y-3)^2)/(49)=1 ?

The foci ((x+2)^2)/(16)+((y-3)^2)/(49)=1 is (-2,3+sqrt(33)),(-2,3-sqrt(33))

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foci ((x+2)^2)/(16)+((y-3)^2)/(49)=1

foci

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

(- 2, 3 + 33), (- 2, 3 - 33)

Solution steps

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

Calculate ellipse properties

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (- 2, 3), b = 7, a = 4

(- 2, 3 + c), (- 2, 3 - c)

Compute

c :

c = 7^{2} - 4^{2} : 33

(- 2, 3 + 33), (- 2, 3 - 33)

\frac{( x - 6 ) ^{2}}{9} + \frac{( y - 9 ) ^{2}}{64} = 1

y^{2} = 16 - x^{2}

\frac{( x - 11 ) ^{2}}{63} + \frac{( y - 13 ) ^{2}}{81} = 1

\frac{( x - 7 ) ^{2}}{144} + \frac{( y - 4 ) ^{2}}{128} = 1

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

What is the foci ((x+2)^2)/(16)+((y-3)^2)/(49)=1 ?
The foci ((x+2)^2)/(16)+((y-3)^2)/(49)=1 is (-2,3+sqrt(33)),(-2,3-sqrt(33))