What is the foci ((x-4)^2}{25}+\frac{(y-2)^2)/1 =1 ?

The foci ((x-4)^2}{25}+\frac{(y-2)^2)/1 =1 is (4+2sqrt(6),2),(4-2sqrt(6),2)

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

foci

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

(4 + 26, 2), (4 - 26, 2)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (4, 2), a = 5, b = 1

(4 + c, 2), (4 - c, 2)

Compute

c :

c = 5^{2} - 1^{2} : 26

(4 + 26, 2), (4 - 26, 2)

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

\frac{x ^{2}}{289} + \frac{y ^{2}}{225} = 1

4 x^{2} + y^{2} - 32 x + 4 y + 64 = 0

\frac{( x ) ^{2}}{144} + \frac{( y ) ^{2}}{169} = 1

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 1 ) ^{2}}{9} = 1

What is the foci ((x-4)^2}{25}+\frac{(y-2)^2)/1 =1 ?
The foci ((x-4)^2}{25}+\frac{(y-2)^2)/1 =1 is (4+2sqrt(6),2),(4-2sqrt(6),2)