What is the foci ((x+4)^2)/(16)+((y-1)^2)/(36)=1 ?

The foci ((x+4)^2)/(16)+((y-1)^2)/(36)=1 is (-4,1+2sqrt(5)),(-4,1-2sqrt(5))

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foci ((x+4)^2)/(16)+((y-1)^2)/(36)=1

foci

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 1 ) ^{2}}{36} = 1

(- 4, 1 + 25), (- 4, 1 - 25)

Solution steps

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

Calculate ellipse properties

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 1 ) ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (- 4, 1), b = 6, a = 4

(- 4, 1 + c), (- 4, 1 - c)

Compute

c :

c = 6^{2} - 4^{2} : 25

(- 4, 1 + 25), (- 4, 1 - 25)

16 x^{2} + 25 y^{2} + 100 y - 300 = 0

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{4 ^{2}} = 1

x^{2} + 3 y^{2} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{36} = 1

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

What is the foci ((x+4)^2)/(16)+((y-1)^2)/(36)=1 ?
The foci ((x+4)^2)/(16)+((y-1)^2)/(36)=1 is (-4,1+2sqrt(5)),(-4,1-2sqrt(5))