What is the foci ((x-3)^2}{25}+\frac{y^2)/9 =1 ?

The foci ((x-3)^2}{25}+\frac{y^2)/9 =1 is (7,0),(-1,0)

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foci ((x-3)^2}{25}+\frac{y^2)/9 =1

foci

\frac{( x - 3 ) ^{2}}{25} + \frac{y ^{2}}{9} = 1

(7, 0), (- 1, 0)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{25} + \frac{y ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (3, 0), a = 5, b = 3

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

Compute

c :

c = 5^{2} - 3^{2} : 4

(3 + 4, 0), (3 - 4, 0)

Refine

(7, 0), (- 1, 0)

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

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1

25 x^{2} - 200 x + y^{2} + 4 y + 379 = 0

64 x^{2} + 100 y^{2} + 768 x - 600 y - 3196 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

What is the foci ((x-3)^2}{25}+\frac{y^2)/9 =1 ?
The foci ((x-3)^2}{25}+\frac{y^2)/9 =1 is (7,0),(-1,0)