What is the foci-9x^2+4y^2=36 ?

The foci-9x^2+4y^2=36 is (0,sqrt(13)),(0,-sqrt(13))

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foci-9x^2+4y^2=36

foci

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

(0, 13), (0, - 13)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (0, 0), a = 3, b = 2

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

Compute

c :

c = 3^{2} + 2^{2} : 13

(0, 0 + 13), (0, 0 - 13)

Refine

(0, 13), (0, - 13)

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

- \frac{x ^{2}}{16} + \frac{y ^{2}}{36} = 1

\frac{y ^{2}}{25} - \frac{( x - 2 ) ^{2}}{0} = 1

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

\frac{x ^{2}}{25} - \frac{y ^{2}}{144} = 1

What is the foci-9x^2+4y^2=36 ?
The foci-9x^2+4y^2=36 is (0,sqrt(13)),(0,-sqrt(13))