What is the foci ((x-3)^2)/(64)+((y+1)^2)/(49)=1 ?

The foci ((x-3)^2)/(64)+((y+1)^2)/(49)=1 is (3+sqrt(15),-1),(3-sqrt(15),-1)

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foci ((x-3)^2)/(64)+((y+1)^2)/(49)=1

foci

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 1 ) ^{2}}{49} = 1

(3 + 15, - 1), (3 - 15, - 1)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 1 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (3, - 1), a = 8, b = 7

(3 + c, - 1), (3 - c, - 1)

Compute

c :

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

(3 + 15, - 1), (3 - 15, - 1)

\frac{( x + 20 ) ^{2}}{16} + \frac{y ^{2}}{89} = 1

\frac{x ^{2}}{\frac{9}{4}} + \frac{y ^{2}}{\frac{25}{4}} = 1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{4} = 1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{4} = 1

\frac{( x - 8 ) ^{2}}{2} + \frac{( y - 3 ) ^{2}}{27} = 1

What is the foci ((x-3)^2)/(64)+((y+1)^2)/(49)=1 ?
The foci ((x-3)^2)/(64)+((y+1)^2)/(49)=1 is (3+sqrt(15),-1),(3-sqrt(15),-1)