What is the foci ((x-3)^2)/(25)+((y-4)^2)/(16)=1 ?

The foci ((x-3)^2)/(25)+((y-4)^2)/(16)=1 is (6,4),(0,4)

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foci ((x-3)^2)/(25)+((y-4)^2)/(16)=1

foci

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{16} = 1

(6, 4), (0, 4)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{16} = 1 :

Ellipse with center

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

(3 + c, 4), (3 - c, 4)

Compute

c :

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

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

Refine

(6, 4), (0, 4)

\frac{1 x ^{2}}{16} + \frac{y ^{2}}{9} = 1

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{49} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 5 ) ^{2}}{4} = 1

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

What is the foci ((x-3)^2)/(25)+((y-4)^2)/(16)=1 ?
The foci ((x-3)^2)/(25)+((y-4)^2)/(16)=1 is (6,4),(0,4)