What is the foci ((x-4)^2)/9+((y-7)^2)/(36)=1 ?

The foci ((x-4)^2)/9+((y-7)^2)/(36)=1 is (4,7+3sqrt(3)),(4,7-3sqrt(3))

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foci ((x-4)^2)/9+((y-7)^2)/(36)=1

foci

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 7 ) ^{2}}{36} = 1

(4, 7 + 33), (4, 7 - 33)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 7 ) ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (4, 7), b = 6, a = 3

(4, 7 + c), (4, 7 - c)

Compute

c :

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

(4, 7 + 33), (4, 7 - 33)

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

x^{2} + 4 y^{2} = 36

81 x^{2} + 9 y^{2} = 9

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

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{121} = 1

What is the foci ((x-4)^2)/9+((y-7)^2)/(36)=1 ?
The foci ((x-4)^2)/9+((y-7)^2)/(36)=1 is (4,7+3sqrt(3)),(4,7-3sqrt(3))