What is the foci ((x+6)^2)/7+((y-3)^2)/(16)=1 ?

The foci ((x+6)^2)/7+((y-3)^2)/(16)=1 is (-6,6),(-6,0)

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foci ((x+6)^2)/7+((y-3)^2)/(16)=1

foci

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

(- 6, 6), (- 6, 0)

Solution steps

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

Calculate ellipse properties

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (- 6, 3), b = 4, a = 7

(- 6, 3 + c), (- 6, 3 - c)

Compute

c :

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

(- 6, 3 + 3), (- 6, 3 - 3)

Refine

(- 6, 6), (- 6, 0)

\frac{( x - 4 ) ^{2}}{3} + \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{36} + \frac{8 y ^{2}}{81} = 1

x^{2} + 3 y^{2} + 8 x + 9 y - 14 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{13} = 1

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1

What is the foci ((x+6)^2)/7+((y-3)^2)/(16)=1 ?
The foci ((x+6)^2)/7+((y-3)^2)/(16)=1 is (-6,6),(-6,0)