What is the foci ((x-6)^2)/9+((y-9)^2)/(64)=1 ?

The foci ((x-6)^2)/9+((y-9)^2)/(64)=1 is (6,9+sqrt(55)),(6,9-sqrt(55))

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foci ((x-6)^2)/9+((y-9)^2)/(64)=1

foci

\frac{( x - 6 ) ^{2}}{9} + \frac{( y - 9 ) ^{2}}{64} = 1

(6, 9 + 55), (6, 9 - 55)

Solution steps

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

Calculate ellipse properties

\frac{( x - 6 ) ^{2}}{9} + \frac{( y - 9 ) ^{2}}{64} = 1 :

Ellipse with center

(h, k) = (6, 9), b = 8, a = 3

(6, 9 + c), (6, 9 - c)

Compute

c :

c = 8^{2} - 3^{2} : 55

(6, 9 + 55), (6, 9 - 55)

y^{2} = 16 - x^{2}

\frac{( x - 11 ) ^{2}}{63} + \frac{( y - 13 ) ^{2}}{81} = 1

\frac{( x - 7 ) ^{2}}{144} + \frac{( y - 4 ) ^{2}}{128} = 1

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

37 x^{2} + y^{2} = 37

What is the foci ((x-6)^2)/9+((y-9)^2)/(64)=1 ?
The foci ((x-6)^2)/9+((y-9)^2)/(64)=1 is (6,9+sqrt(55)),(6,9-sqrt(55))