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

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

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

foci

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

(- 2, 33), (- 2, - 33)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, 0), b = 7, a = 4

(- 2, 0 + c), (- 2, 0 - c)

Compute

c :

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

(- 2, 0 + 33), (- 2, 0 - 33)

Refine

(- 2, 33), (- 2, - 33)

16 x^{2} + 192 x + 576 + 25 y^{2} = 1600

\frac{( x - 2 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{9} = 1

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

5 x^{2} + 9 y^{2} + 30 x = 0

\frac{x ^{2}}{64} + \frac{y ^{2}}{9} = 1

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