What is the foci ((x-5)^2}{25}+\frac{(y-3)^2)/9 =1 ?

The foci ((x-5)^2}{25}+\frac{(y-3)^2)/9 =1 is (9,3),(1,3)

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foci ((x-5)^2}{25}+\frac{(y-3)^2)/9 =1

foci

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{9} = 1

(9, 3), (1, 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{9} = 1 :

Ellipse with center

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

(5 + c, 3), (5 - c, 3)

Compute

c :

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

(5 + 4, 3), (5 - 4, 3)

Refine

(9, 3), (1, 3)

\frac{( x + 1 ) ^{2}}{25} + \frac{y ^{2}}{16} = 1

(\frac{x}{2.5})^{2} + (\frac{y}{3})^{2} = 1

9 x^{2} + 4 y^{2} + 54 x - 8 y + 49 = 0

25 (x - 1)^{2} + 9 (y + 4)^{2} = 225

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

What is the foci ((x-5)^2}{25}+\frac{(y-3)^2)/9 =1 ?
The foci ((x-5)^2}{25}+\frac{(y-3)^2)/9 =1 is (9,3),(1,3)