What is the foci ((x-8)^2}{49}+\frac{(y+5)^2)/9 =1 ?

The foci ((x-8)^2}{49}+\frac{(y+5)^2)/9 =1 is (8+2sqrt(10),-5),(8-2sqrt(10),-5)

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foci ((x-8)^2}{49}+\frac{(y+5)^2)/9 =1

foci

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{9} = 1

(8 + 210, - 5), (8 - 210, - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (8, - 5), a = 7, b = 3

(8 + c, - 5), (8 - c, - 5)

Compute

c :

c = 7^{2} - 3^{2} : 210

(8 + 210, - 5), (8 - 210, - 5)

\frac{( x - 4 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{36} = 1

\frac{( x + 1 ) ^{2}}{28} + \frac{( y + 5 ) ^{2}}{23} = 1

x^{2} - 8 x + 25 y^{2} - 100 y + 91 = 0

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

What is the foci ((x-8)^2}{49}+\frac{(y+5)^2)/9 =1 ?
The foci ((x-8)^2}{49}+\frac{(y+5)^2)/9 =1 is (8+2sqrt(10),-5),(8-2sqrt(10),-5)