What is the foci ((x-3)^2}{16}+\frac{(y+4)^2)/9 =1 ?

The foci ((x-3)^2}{16}+\frac{(y+4)^2)/9 =1 is (3+sqrt(7),-4),(3-sqrt(7),-4)

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foci ((x-3)^2}{16}+\frac{(y+4)^2)/9 =1

foci

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

(3 + 7, - 4), (3 - 7, - 4)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1 :

Ellipse with center

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

(3 + c, - 4), (3 - c, - 4)

Compute

c :

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

(3 + 7, - 4), (3 - 7, - 4)

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

4 x^{2} + 25 y^{2} - 50 y = 75

x^{2} + 4 y^{2} - 16 = 0

16 (x - 1)^{2} + 9 (y + 2)^{2} = 144

\frac{( x - 5 ) ^{2}}{3} + \frac{( y - 7 ) ^{2}}{6} = 1

What is the foci ((x-3)^2}{16}+\frac{(y+4)^2)/9 =1 ?
The foci ((x-3)^2}{16}+\frac{(y+4)^2)/9 =1 is (3+sqrt(7),-4),(3-sqrt(7),-4)