What is the foci (x/5)^2+(y/3)^2=1 ?

The foci (x/5)^2+(y/3)^2=1 is (4,0),(-4,0)

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foci (x/5)^2+(y/3)^2=1

foci

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

(4, 0), (- 4, 0)

Solution steps

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

Calculate ellipse properties

(\frac{x}{5})^{2} + (\frac{y}{3})^{2} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 5, b = 3

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

Compute

c :

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

(0 + 4, 0), (0 - 4, 0)

Refine

(4, 0), (- 4, 0)

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

4 (x - 4)^{2} + 9 y^{2} = 36

\frac{( x - 1 ) ^{2}}{45} + \frac{( y - 7 ) ^{2}}{81} = 1

\frac{x ^{2}}{67} + \frac{y ^{2}}{57} = 1

4 x^{2} + 16 y^{2} = 64