What is the foci ((x-3)^2}{16}+\frac{(y+1)^2)/9 =1 ?

The foci ((x-3)^2}{16}+\frac{(y+1)^2)/9 =1 is (3+sqrt(7),-1),(3-sqrt(7),-1)

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foci ((x-3)^2}{16}+\frac{(y+1)^2)/9 =1

foci

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{9} = 1

(3 + 7, - 1), (3 - 7, - 1)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (3, - 1), a = 4, b = 3

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

Compute

c :

c = 4^{2} - 3^{2} : 7

(3 + 7, - 1), (3 - 7, - 1)

5 x^{2} + 2 y^{2} = 10

\frac{x ^{2}}{36} + \frac{y ^{2}}{81} = 1

\frac{y ^{2}}{81} + \frac{x ^{2}}{36} = 1

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 1 ) ^{2}}{36} = 1

16 x^{2} + 25 y^{2} + 100 y - 300 = 0

What is the foci ((x-3)^2}{16}+\frac{(y+1)^2)/9 =1 ?
The foci ((x-3)^2}{16}+\frac{(y+1)^2)/9 =1 is (3+sqrt(7),-1),(3-sqrt(7),-1)