What is the foci ((y-6)^2)/(36)-((x+1)^2)/(16)=1 ?

The foci ((y-6)^2)/(36)-((x+1)^2)/(16)=1 is (-1,6+2sqrt(13)),(-1,6-2sqrt(13))

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foci ((y-6)^2)/(36)-((x+1)^2)/(16)=1

foci

\frac{( y - 6 ) ^{2}}{36} - \frac{( x + 1 ) ^{2}}{16} = 1

(- 1, 6 + 213), (- 1, 6 - 213)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 6 ) ^{2}}{36} - \frac{( x + 1 ) ^{2}}{16} = 1 :

Up-down Hyperbola with

(h, k) = (- 1, 6), a = 6, b = 4

(- 1, 6 + c), (- 1, 6 - c)

Compute

c :

c = 6^{2} + 4^{2} : 213

(- 1, 6 + 213), (- 1, 6 - 213)

9 x^{2} - 4 y^{2} + 54 x + 16 y + 29 = 0

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

\frac{x ^{2}}{49} - \frac{y ^{2}}{4} = 1

\frac{( y + \frac{7}{2} ) ^{2}}{25} - \frac{( x - \frac{9}{2} ) ^{2}}{25} = 1

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

What is the foci ((y-6)^2)/(36)-((x+1)^2)/(16)=1 ?
The foci ((y-6)^2)/(36)-((x+1)^2)/(16)=1 is (-1,6+2sqrt(13)),(-1,6-2sqrt(13))