What is the foci ((x-6)^2)/(16)+((y+3)^2)/(25)=1 ?

The foci ((x-6)^2)/(16)+((y+3)^2)/(25)=1 is (6,0),(6,-6)

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foci ((x-6)^2)/(16)+((y+3)^2)/(25)=1

foci

\frac{( x - 6 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x - 6 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (6, - 3), b = 5, a = 4

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

Compute

c :

c = 5^{2} - 4^{2} : 3

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

Refine

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

4 x^{2} + 25 y^{2} + 24 x + 250 y = - 561

4 x^{2} + 36 y^{2} + 24 x + 432 y = - 1188

\frac{( x - 2 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{9} = 1

\frac{x ^{2}}{25} + \frac{y ^{2}}{169} = 1

16 x^{2} - 96 x + 9 y^{2} = 0

What is the foci ((x-6)^2)/(16)+((y+3)^2)/(25)=1 ?
The foci ((x-6)^2)/(16)+((y+3)^2)/(25)=1 is (6,0),(6,-6)