What is the foci (x/(2.5))^2+(y/3)^2=1 ?

The foci (x/(2.5))^2+(y/3)^2=1 is (0,1.65831…),(0,-1.65831…)

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foci (x/(2.5))^2+(y/3)^2=1

foci

(\frac{x}{2.5})^{2} + (\frac{y}{3})^{2} = 1

(0, 1.65831 \dots), (0, - 1.65831 \dots)

Solution steps

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

Calculate ellipse properties

(\frac{x}{2.5})^{2} + (\frac{y}{3})^{2} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 3, a = 2.5

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

Compute

c :

c = 3^{2} - 2. 5^{2} : 1.65831 \dots

(0, 0 + 1.65831 \dots), (0, 0 - 1.65831 \dots)

Refine

(0, 1.65831 \dots), (0, - 1.65831 \dots)

9 x^{2} + 4 y^{2} + 54 x - 8 y + 49 = 0

25 (x - 1)^{2} + 9 (y + 4)^{2} = 225

\frac{( x + 1 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

\frac{( x - 6 ) ^{2}}{9} + \frac{( y - 9 ) ^{2}}{64} = 1

What is the foci (x/(2.5))^2+(y/3)^2=1 ?
The foci (x/(2.5))^2+(y/3)^2=1 is (0,1.65831…),(0,-1.65831…)