What is the foci 4y^2-9x^2+40y-540x-8036=0 ?

The foci 4y^2-9x^2+40y-540x-8036=0 is (-30,-5+sqrt(13)),(-30,-5-sqrt(13))

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foci 4y^2-9x^2+40y-540x-8036=0

foci

4 y^{2} - 9 x^{2} + 40 y - 540 x - 8036 = 0

(- 30, - 5 + 13), (- 30, - 5 - 13)

Solution steps

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

Calculate Hyperbola properties

4 y^{2} - 9 x^{2} + 40 y - 540 x - 8036 = 0 :

Up-down Hyperbola with

(h, k) = (- 30, - 5), a = 3, b = 2

(- 30, - 5 + c), (- 30, - 5 - c)

Compute

c :

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

(- 30, - 5 + 13), (- 30, - 5 - 13)

\frac{x ^{2}}{49} - y^{2} = 1

4 x^{2} - 4 y^{2} + 20 x - 16 y + 25 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{81} = 1

3 x^{2} - 4 y^{2} = 12

5 x^{2} - 5 y^{2} + 20 x - 60 y = 185

What is the foci 4y^2-9x^2+40y-540x-8036=0 ?
The foci 4y^2-9x^2+40y-540x-8036=0 is (-30,-5+sqrt(13)),(-30,-5-sqrt(13))