What is the foci ((x-5)^2)/(12)-((y+6)^2)/(17)=1 ?

The foci ((x-5)^2)/(12)-((y+6)^2)/(17)=1 is (5+sqrt(29),-6),(5-sqrt(29),-6)

Popular >

foci ((x-5)^2)/(12)-((y+6)^2)/(17)=1

foci

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

(5 + 29, - 6), (5 - 29, - 6)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (5, - 6), a = 23, b = 17

(5 + c, - 6), (5 - c, - 6)

Compute

c :

c = (23)^{2} + 17^{2} : 29

(5 + 29, - 6), (5 - 29, - 6)

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

\frac{( x + 2 ) ^{2}}{25} - \frac{( y - 1 ) ^{2}}{169} = 1

\frac{( x - 2 ) ^{2}}{5} - \frac{( y + 1 ) ^{2}}{9} = 1

What is the foci ((x-5)^2)/(12)-((y+6)^2)/(17)=1 ?
The foci ((x-5)^2)/(12)-((y+6)^2)/(17)=1 is (5+sqrt(29),-6),(5-sqrt(29),-6)