What is the foci y^2=1-25x^2 ?

The foci y^2=1-25x^2 is (0,(2sqrt(6))/5),(0,-(2sqrt(6))/5)

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foci y^2=1-25x^2

foci

y^{2} = 1 - 25 x^{2}

(0, \frac{2 6}{5}), (0, - \frac{2 6}{5})

Solution steps

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

Calculate ellipse properties

y^{2} = 1 - 25 x^{2} :

Ellipse with center

(h, k) = (0, 0), b = 1, a = \frac{1}{5}

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

Compute

c :

c = 1^{2} - (\frac{1}{5})^{2} : \frac{2 6}{5}

(0, 0 + \frac{2 6}{5}), (0, 0 - \frac{2 6}{5})

Refine

(0, \frac{2 6}{5}), (0, - \frac{2 6}{5})

2 x^{2} - 8 x + 3 y^{2} + 6 y + 5 = 0

36 x^{2} + 288 x + 64 y^{2} - 256 y = 1472

\frac{x ^{2}}{81} + \frac{y ^{2}}{56} = 1

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

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

What is the foci y^2=1-25x^2 ?
The foci y^2=1-25x^2 is (0,(2sqrt(6))/5),(0,-(2sqrt(6))/5)