What is the foci ((x-3)^2}{11}-\frac{(y+1)^2)/5 =1 ?

The foci ((x-3)^2}{11}-\frac{(y+1)^2)/5 =1 is (7,-1),(-1,-1)

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foci ((x-3)^2}{11}-\frac{(y+1)^2)/5 =1

foci

\frac{( x - 3 ) ^{2}}{11} - \frac{( y + 1 ) ^{2}}{5} = 1

(7, - 1), (- 1, - 1)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{11} - \frac{( y + 1 ) ^{2}}{5} = 1 :

Right-left Hyperbola with

(h, k) = (3, - 1), a = 11, b = 5

(3 + c, - 1), (3 - c, - 1)

Compute

c :

c = 11^{2} + 5^{2} : 4

(3 + 4, - 1), (3 - 4, - 1)

Refine

(7, - 1), (- 1, - 1)

16 x^{2} - y^{2} - 96 x - 6 y + 119 = 0

9 x^{2} - 18 x - 16 y^{2} - 64 y = 199

18 x^{2} - 32 y^{2} = 2

16 x^{2} - 9 y^{2} - 64 x - 18 y + 199 = 0

\frac{x ^{2}}{144} - \frac{y ^{2}}{81} = 1

What is the foci ((x-3)^2}{11}-\frac{(y+1)^2)/5 =1 ?
The foci ((x-3)^2}{11}-\frac{(y+1)^2)/5 =1 is (7,-1),(-1,-1)