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

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

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

foci

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

(3, 0), (- 3, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = 5, b = 4

(0 + c, 0), (0 - c, 0)

Compute

c :

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

(0 + 3, 0), (0 - 3, 0)

Refine

(3, 0), (- 3, 0)

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

4 (x + 2)^{2} + 16 (y - 2)^{2} = 64

\frac{( x - 7 ) ^{2}}{64} + \frac{( y + 2 ) ^{2}}{25} = 1

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

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

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