What is the foci ((x-7)^2)/(64)+((y+2)^2)/(25)=1 ?

The foci ((x-7)^2)/(64)+((y+2)^2)/(25)=1 is (7+sqrt(39),-2),(7-sqrt(39),-2)

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foci ((x-7)^2)/(64)+((y+2)^2)/(25)=1

foci

\frac{( x - 7 ) ^{2}}{64} + \frac{( y + 2 ) ^{2}}{25} = 1

(7 + 39, - 2), (7 - 39, - 2)

Solution steps

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

Calculate ellipse properties

\frac{( x - 7 ) ^{2}}{64} + \frac{( y + 2 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (7, - 2), a = 8, b = 5

(7 + c, - 2), (7 - c, - 2)

Compute

c :

c = 8^{2} - 5^{2} : 39

(7 + 39, - 2), (7 - 39, - 2)

\frac{( x - 4 ) ^{2}}{25} + \frac{( y - 2 ) ^{2}}{1} = 1

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

\frac{x ^{2}}{289} + \frac{y ^{2}}{225} = 1

4 x^{2} + y^{2} - 32 x + 4 y + 64 = 0

\frac{( x ) ^{2}}{144} + \frac{( y ) ^{2}}{169} = 1

What is the foci ((x-7)^2)/(64)+((y+2)^2)/(25)=1 ?
The foci ((x-7)^2)/(64)+((y+2)^2)/(25)=1 is (7+sqrt(39),-2),(7-sqrt(39),-2)