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)

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

x = - 5 y^{2}

9 x^{2} + 16 y^{2} - 72 x + 64 y - 368 = 0

y^{2} = - 16 x

y^{2} = - 16 x

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)