What is the foci ((x+10)^2}{16}+\frac{(y-5)^2)/5 =1 ?

The foci ((x+10)^2}{16}+\frac{(y-5)^2)/5 =1 is (-10+sqrt(11),5),(-10-sqrt(11),5)

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foci ((x+10)^2}{16}+\frac{(y-5)^2)/5 =1

foci

\frac{( x + 10 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{5} = 1

(- 10 + 11, 5), (- 10 - 11, 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 10 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{5} = 1 :

Ellipse with center

(h, k) = (- 10, 5), a = 4, b = 5

(- 10 + c, 5), (- 10 - c, 5)

Compute

c :

c = 4^{2} - 5^{2} : 11

(- 10 + 11, 5), (- 10 - 11, 5)

16 x^{2} + 36 y^{2} = 576

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

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

\frac{( x + 5 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{4} = 1

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{3} = 1

What is the foci ((x+10)^2}{16}+\frac{(y-5)^2)/5 =1 ?
The foci ((x+10)^2}{16}+\frac{(y-5)^2)/5 =1 is (-10+sqrt(11),5),(-10-sqrt(11),5)