What is the foci ((x-4)^2)/9+((y-1)^2)/(25)=1 ?

The foci ((x-4)^2)/9+((y-1)^2)/(25)=1 is (4,5),(4,-3)

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foci ((x-4)^2)/9+((y-1)^2)/(25)=1

foci

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

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

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (4, 1), b = 5, a = 3

(4, 1 + c), (4, 1 - c)

Compute

c :

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

(4, 1 + 4), (4, 1 - 4)

Refine

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

\frac{x ^{2}}{81} + \frac{y ^{2}}{100} = 1

x^{2} + 4 y^{2} - 6 x + 16 y + 21 = 0

\frac{x ^{2}}{64} + \frac{y ^{2}}{15} = 1

4 x^{2} + 24 x + 25 y^{2} + 200 y + 336 = 0

\frac{( x + 1 ) ^{2}}{10} + \frac{( y - 3 ) ^{2}}{9} = 1

What is the foci ((x-4)^2)/9+((y-1)^2)/(25)=1 ?
The foci ((x-4)^2)/9+((y-1)^2)/(25)=1 is (4,5),(4,-3)