What is the foci ((y+6)^2)/(36)-((x+4)^2)/(48)=1 ?

The foci ((y+6)^2)/(36)-((x+4)^2)/(48)=1 is (-4,-6+2sqrt(21)),(-4,-6-2sqrt(21))

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foci ((y+6)^2)/(36)-((x+4)^2)/(48)=1

foci

\frac{( y + 6 ) ^{2}}{36} - \frac{( x + 4 ) ^{2}}{48} = 1

(- 4, - 6 + 221), (- 4, - 6 - 221)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 6 ) ^{2}}{36} - \frac{( x + 4 ) ^{2}}{48} = 1 :

Up-down Hyperbola with

(h, k) = (- 4, - 6), a = 6, b = 43

(- 4, - 6 + c), (- 4, - 6 - c)

Compute

c :

c = 6^{2} + (43)^{2} : 221

(- 4, - 6 + 221), (- 4, - 6 - 221)

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

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

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

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

\frac{( x ) ^{2}}{\frac{25}{4}} - \frac{( y ) ^{2}}{\frac{11}{4}} = 1

What is the foci ((y+6)^2)/(36)-((x+4)^2)/(48)=1 ?
The foci ((y+6)^2)/(36)-((x+4)^2)/(48)=1 is (-4,-6+2sqrt(21)),(-4,-6-2sqrt(21))