What is the foci ((x-6)^2)/(10)-((y+2)^2)/(17)=1 ?

The foci ((x-6)^2)/(10)-((y+2)^2)/(17)=1 is (6+3sqrt(3),-2),(6-3sqrt(3),-2)

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foci ((x-6)^2)/(10)-((y+2)^2)/(17)=1

foci

\frac{( x - 6 ) ^{2}}{10} - \frac{( y + 2 ) ^{2}}{17} = 1

(6 + 33, - 2), (6 - 33, - 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 6 ) ^{2}}{10} - \frac{( y + 2 ) ^{2}}{17} = 1 :

Right-left Hyperbola with

(h, k) = (6, - 2), a = 10, b = 17

(6 + c, - 2), (6 - c, - 2)

Compute

c :

c = 10^{2} + 17^{2} : 33

(6 + 33, - 2), (6 - 33, - 2)

\frac{( x - 1 ) ^{2}}{1 2 ^{2}} - \frac{( y - 11 ) ^{2}}{8 ^{2}} = 1

\frac{( x + 1 ) ^{2}}{4} - \frac{( y - 2 ) ^{2}}{12} = 1

\frac{( y - 3 ) ^{2}}{7} - \frac{( x + 6 ) ^{2}}{42} = 1

\frac{x ^{2}}{3} - \frac{y ^{2}}{3} = 1

2 x^{2} - 3 y^{2} = 6

What is the foci ((x-6)^2)/(10)-((y+2)^2)/(17)=1 ?
The foci ((x-6)^2)/(10)-((y+2)^2)/(17)=1 is (6+3sqrt(3),-2),(6-3sqrt(3),-2)