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

The foci ((x-2)^2}{16}+\frac{y^2)/9 =1 is (2+sqrt(7),0),(2-sqrt(7),0)

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

foci

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

(2 + 7, 0), (2 - 7, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, 0), a = 4, b = 3

(2 + c, 0), (2 - c, 0)

Compute

c :

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

(2 + 7, 0), (2 - 7, 0)

\frac{x ^{2}}{8} + \frac{y ^{2}}{5} = 1

16 x^{2} + 25 y^{2} - 96 x + 100 y - 156 = 0

+ \frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{25} = 1

81 x^{2} + 64 y^{2} = 5184

What is the foci ((x-2)^2}{16}+\frac{y^2)/9 =1 ?
The foci ((x-2)^2}{16}+\frac{y^2)/9 =1 is (2+sqrt(7),0),(2-sqrt(7),0)