What is the foci ((x-12)^2)/(289)+((y-3)^2)/(64)=1 ?

The foci ((x-12)^2)/(289)+((y-3)^2)/(64)=1 is (27,3),(-3,3)

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foci ((x-12)^2)/(289)+((y-3)^2)/(64)=1

foci

\frac{( x - 12 ) ^{2}}{289} + \frac{( y - 3 ) ^{2}}{64} = 1

(27, 3), (- 3, 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 12 ) ^{2}}{289} + \frac{( y - 3 ) ^{2}}{64} = 1 :

Ellipse with center

(h, k) = (12, 3), a = 17, b = 8

(12 + c, 3), (12 - c, 3)

Compute

c :

c = 1 7^{2} - 8^{2} : 15

(12 + 15, 3), (12 - 15, 3)

Refine

(27, 3), (- 3, 3)

9 X^{2} + 4 Y^{2} = 36

\frac{x ^{2}}{9} + \frac{y ^{2}}{36} = 1

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

\frac{( x + 1 ) ^{2}}{100} + \frac{( y + 2 ) ^{2}}{121} = 1

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

What is the foci ((x-12)^2)/(289)+((y-3)^2)/(64)=1 ?
The foci ((x-12)^2)/(289)+((y-3)^2)/(64)=1 is (27,3),(-3,3)