What is the foci ((x-3)^2)/(16)-((y+8)^2)/(64)=1 ?

The foci ((x-3)^2)/(16)-((y+8)^2)/(64)=1 is (3+4sqrt(5),-8),(3-4sqrt(5),-8)

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foci ((x-3)^2)/(16)-((y+8)^2)/(64)=1

foci

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 8 ) ^{2}}{64} = 1

(3 + 45, - 8), (3 - 45, - 8)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 8 ) ^{2}}{64} = 1 :

Right-left Hyperbola with

(h, k) = (3, - 8), a = 4, b = 8

(3 + c, - 8), (3 - c, - 8)

Compute

c :

c = 4^{2} + 8^{2} : 45

(3 + 45, - 8), (3 - 45, - 8)

9 x^{2} - 4 y^{2} - 18 x - 16 y + 29 = 0

9 y^{2} + 36 y - 4 x^{2} - 24 x - 36 = 0

4 (x - 3)^{2} - 9 (y - 2)^{2} = 36

9 x^{2} - y^{2} - 36 x - 4 y + 23 = 0

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

What is the foci ((x-3)^2)/(16)-((y+8)^2)/(64)=1 ?
The foci ((x-3)^2)/(16)-((y+8)^2)/(64)=1 is (3+4sqrt(5),-8),(3-4sqrt(5),-8)