What is the foci ((x8)^2)/(49)+((y+1)^2)/(100)=1 ?

The foci ((x8)^2)/(49)+((y+1)^2)/(100)=1 is (0,(-8+sqrt(6351))/8),(0,(-8-sqrt(6351))/8)

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foci ((x8)^2)/(49)+((y+1)^2)/(100)=1

foci

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

(0, \frac{- 8 + 6351}{8}), (0, \frac{- 8 - 6351}{8})

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) = (0, - 1), b = 10, a = \frac{7}{8}

(0, - 1 + c), (0, - 1 - c)

Compute

c :

c = 1 0^{2} - (\frac{7}{8})^{2} : \frac{6351}{8}

(0, - 1 + \frac{6351}{8}), (0, - 1 - \frac{6351}{8})

Refine

(0, \frac{- 8 + 6351}{8}), (0, \frac{- 8 - 6351}{8})

\frac{x ^{2}}{32} + \frac{y ^{2}}{18} = 1

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

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

y^{2} = 1 - 49 x^{2}

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

What is the foci ((x8)^2)/(49)+((y+1)^2)/(100)=1 ?
The foci ((x8)^2)/(49)+((y+1)^2)/(100)=1 is (0,(-8+sqrt(6351))/8),(0,(-8-sqrt(6351))/8)