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

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

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

foci

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

(2, 0), (- 2, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = \frac{5}{2}, b = \frac{3}{2}

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

Compute

c :

c = (\frac{5}{2})^{2} - (\frac{3}{2})^{2} : 2

(0 + 2, 0), (0 - 2, 0)

Refine

(2, 0), (- 2, 0)

x^{2} + 4 y^{2} - 6 x + 8 y - 3 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{121} = 1

2 x^{2} + y^{2} + 12 x - 2 y + 11 = 0

16 x^{2} + 9 y^{2} + 64 x - 80 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1

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