What is the foci ((x-6)^2)/(36)+((y+4)^2)/(16)=1 ?

The foci ((x-6)^2)/(36)+((y+4)^2)/(16)=1 is (6+2sqrt(5),-4),(6-2sqrt(5),-4)

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foci ((x-6)^2)/(36)+((y+4)^2)/(16)=1

foci

(6 + 25, - 4), (6 - 25, - 4)

Solution steps

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

Calculate ellipse properties

\frac{( x - 6 ) ^{2}}{36} + \frac{( y + 4 ) ^{2}}{16} = 1 :

Ellipse with center

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

(6 + c, - 4), (6 - c, - 4)

Compute

c :

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

(6 + 25, - 4), (6 - 25, - 4)

What is the foci ((x-6)^2)/(36)+((y+4)^2)/(16)=1 ?
The foci ((x-6)^2)/(36)+((y+4)^2)/(16)=1 is (6+2sqrt(5),-4),(6-2sqrt(5),-4)