What is the foci ((x-2)^2}{36}+\frac{(y+5)^2)/9 =1 ?

The foci ((x-2)^2}{36}+\frac{(y+5)^2)/9 =1 is (2+3sqrt(3),-5),(2-3sqrt(3),-5)

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foci ((x-2)^2}{36}+\frac{(y+5)^2)/9 =1

foci

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1

(2 + 33, - 5), (2 - 33, - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (2, - 5), a = 6, b = 3

(2 + c, - 5), (2 - c, - 5)

Compute

c :

c = 6^{2} - 3^{2} : 33

(2 + 33, - 5), (2 - 33, - 5)

25 x^{2} - 200 x + y^{2} + 4 y + 379 = 0

64 x^{2} + 100 y^{2} + 768 x - 600 y - 3196 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

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

3 x^{2} + y^{2} - 24 x + 39 = 0

What is the foci ((x-2)^2}{36}+\frac{(y+5)^2)/9 =1 ?
The foci ((x-2)^2}{36}+\frac{(y+5)^2)/9 =1 is (2+3sqrt(3),-5),(2-3sqrt(3),-5)