What is the foci ((x+3)^2)/(49)+(y^2)/(36)=1 ?

The foci ((x+3)^2)/(49)+(y^2)/(36)=1 is (-3+sqrt(13),0),(-3-sqrt(13),0)

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foci ((x+3)^2)/(49)+(y^2)/(36)=1

foci

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

(- 3 + 13, 0), (- 3 - 13, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 3, 0), a = 7, b = 6

(- 3 + c, 0), (- 3 - c, 0)

Compute

c :

c = 7^{2} - 6^{2} : 13

(- 3 + 13, 0), (- 3 - 13, 0)

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

\frac{( x + 5 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{64} = 1

9 x^{2} + 16 y^{2} - 72 x + 64 y - 368 = 0

7 x^{2} + 2 y^{2} = 14

25 x^{2} + 36 y^{2} = 900

What is the foci ((x+3)^2)/(49)+(y^2)/(36)=1 ?
The foci ((x+3)^2)/(49)+(y^2)/(36)=1 is (-3+sqrt(13),0),(-3-sqrt(13),0)