What is the foci ((x-4)^2)/(32)+((y-3)^2)/(36)=1 ?

The foci ((x-4)^2)/(32)+((y-3)^2)/(36)=1 is (4,5),(4,1)

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foci ((x-4)^2)/(32)+((y-3)^2)/(36)=1

foci

(4, 5), (4, 1)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (4, 3), b = 6, a = 42

(4, 3 + c), (4, 3 - c)

Compute

c :

c = 6^{2} - (42)^{2} : 2

(4, 3 + 2), (4, 3 - 2)

Refine

(4, 5), (4, 1)

What is the foci ((x-4)^2)/(32)+((y-3)^2)/(36)=1 ?
The foci ((x-4)^2)/(32)+((y-3)^2)/(36)=1 is (4,5),(4,1)