What is the foci 9x^2+y^2=1 ?

The foci 9x^2+y^2=1 is (0,(2sqrt(2))/3),(0,-(2sqrt(2))/3)

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foci 9x^2+y^2=1

foci

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

(0, \frac{2 2}{3}), (0, - \frac{2 2}{3})

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 1, a = \frac{1}{3}

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

Compute

c :

c = 1^{2} - (\frac{1}{3})^{2} : \frac{2 2}{3}

(0, 0 + \frac{2 2}{3}), (0, 0 - \frac{2 2}{3})

Refine

(0, \frac{2 2}{3}), (0, - \frac{2 2}{3})

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

9 (x - 5)^{2} + 4 (y + 2)^{2} = 36

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

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

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

What is the foci 9x^2+y^2=1 ?
The foci 9x^2+y^2=1 is (0,(2sqrt(2))/3),(0,-(2sqrt(2))/3)