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

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

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

foci

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

(2 + 17, - 3), (2 - 17, - 3)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (2, - 3), a = 4, b = 1

(2 + c, - 3), (2 - c, - 3)

Compute

c :

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

(2 + 17, - 3), (2 - 17, - 3)

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

\frac{( x + 2 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{169} = 1

y^{2} - 49 x^{2} - 14 y - 490 x - 1125 = 0

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

4 y^{2} - x^{2} + 6 x + 8 y + 4 = 0

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