What is the foci ((x-6)^2}{25}-\frac{(y+4)^2)/9 =1 ?

The foci ((x-6)^2}{25}-\frac{(y+4)^2)/9 =1 is (6+sqrt(34),-4),(6-sqrt(34),-4)

Popular >

foci ((x-6)^2}{25}-\frac{(y+4)^2)/9 =1

foci

\frac{( x - 6 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{9} = 1

(6 + 34, - 4), (6 - 34, - 4)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 6 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (6, - 4), a = 5, b = 3

(6 + c, - 4), (6 - c, - 4)

Compute

c :

c = 5^{2} + 3^{2} : 34

(6 + 34, - 4), (6 - 34, - 4)

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

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

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

\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

What is the foci ((x-6)^2}{25}-\frac{(y+4)^2)/9 =1 ?
The foci ((x-6)^2}{25}-\frac{(y+4)^2)/9 =1 is (6+sqrt(34),-4),(6-sqrt(34),-4)