What is the foci ((x-4)^2)/(25)+((y-2)^2)/(16)=1 ?

The foci ((x-4)^2)/(25)+((y-2)^2)/(16)=1 is (7,2),(1,2)

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foci ((x-4)^2)/(25)+((y-2)^2)/(16)=1

foci

\frac{( x - 4 ) ^{2}}{25} + \frac{( y - 2 ) ^{2}}{16} = 1

(7, 2), (1, 2)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{25} + \frac{( y - 2 ) ^{2}}{16} = 1 :

Ellipse with center

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

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

Compute

c :

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

(4 + 3, 2), (4 - 3, 2)

Refine

(7, 2), (1, 2)

14 x^{2} + 5 y^{2} - 70 = 0

\frac{( x - 4 ) ^{2}}{5 ^{2}} + \frac{( y - 1 ) ^{2}}{2 ^{2}} = 1

16 x^{2} + 25 y^{2} + 100 y = 300

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

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

What is the foci ((x-4)^2)/(25)+((y-2)^2)/(16)=1 ?
The foci ((x-4)^2)/(25)+((y-2)^2)/(16)=1 is (7,2),(1,2)