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

The foci ((x-3)^2}{16}+\frac{y^2)/9 =1 is (3+sqrt(7),0),(3-sqrt(7),0)

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

foci

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

(3 + 7, 0), (3 - 7, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(3 + c, 0), (3 - c, 0)

Compute

c :

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

(3 + 7, 0), (3 - 7, 0)

\frac{( x + 8 ) ^{2}}{169} + \frac{( y + 9 ) ^{2}}{25} = 1

9 x^{2} + 25 y^{2} + 36 x + 250 y = - 436

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

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

x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0

What is the foci ((x-3)^2}{16}+\frac{y^2)/9 =1 ?
The foci ((x-3)^2}{16}+\frac{y^2)/9 =1 is (3+sqrt(7),0),(3-sqrt(7),0)