What is the foci ((x-8)^2)/(49)+((y+1)^2)/(100)=1 ?

The foci ((x-8)^2)/(49)+((y+1)^2)/(100)=1 is (8,-1+sqrt(51)),(8,-1-sqrt(51))

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foci ((x-8)^2)/(49)+((y+1)^2)/(100)=1

foci

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 1 ) ^{2}}{100} = 1

(8, - 1 + 51), (8, - 1 - 51)

Solution steps

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

Calculate ellipse properties

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 1 ) ^{2}}{100} = 1 :

Ellipse with center

(h, k) = (8, - 1), b = 10, a = 7

(8, - 1 + c), (8, - 1 - c)

Compute

c :

c = 1 0^{2} - 7^{2} : 51

(8, - 1 + 51), (8, - 1 - 51)

\frac{( x - 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{25} = 1

36 x^{2} + 9 y^{2} + 36 y - 288 = 0

100 x^{2} + 36 y^{2} = 225

\frac{y ^{2}}{16} + \frac{x ^{2}}{9} = 1

x^{2} + 3 y^{2} - 6 x + 6 y = 0

What is the foci ((x-8)^2)/(49)+((y+1)^2)/(100)=1 ?
The foci ((x-8)^2)/(49)+((y+1)^2)/(100)=1 is (8,-1+sqrt(51)),(8,-1-sqrt(51))