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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

\frac{y ^{2}}{16} + \frac{x ^{2}}{25} = 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{1}{4} (x - 5)^{2} + \frac{1}{9} (y + 5)^{2} = 1

\frac{( x + 1 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{49} = 1

4 x^{2} + 36 y^{2} = 144

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

1 x^{2} + 4 y^{2} + 10 x + 9 y - 10 = 0

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