What is the foci ((x-3)^2}{16}-\frac{(y+4)^2)/1 =1 ?

The foci ((x-3)^2}{16}-\frac{(y+4)^2)/1 =1 is (3+sqrt(17),-4),(3-sqrt(17),-4)

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foci ((x-3)^2}{16}-\frac{(y+4)^2)/1 =1

foci

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 4 ) ^{2}}{1} = 1

(3 + 17, - 4), (3 - 17, - 4)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 4 ) ^{2}}{1} = 1 :

Right-left Hyperbola with

(h, k) = (3, - 4), a = 4, b = 1

(3 + c, - 4), (3 - c, - 4)

Compute

c :

c = 4^{2} + 1^{2} : 17

(3 + 17, - 4), (3 - 17, - 4)

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

\frac{( x - 5 ) ^{2}}{12} - \frac{( y + 6 ) ^{2}}{17} = 1

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

\frac{y ^{2}}{49} - \frac{x ^{2}}{16} = 1

25 x^{2} - 16 y^{2} + 250 x + 32 y + 109 = 0

What is the foci ((x-3)^2}{16}-\frac{(y+4)^2)/1 =1 ?
The foci ((x-3)^2}{16}-\frac{(y+4)^2)/1 =1 is (3+sqrt(17),-4),(3-sqrt(17),-4)