What is the foci ((x-3)^2)/9-((y-2)^2)/(22)=1 ?

The foci ((x-3)^2)/9-((y-2)^2)/(22)=1 is (3+sqrt(31),2),(3-sqrt(31),2)

Popular >

foci ((x-3)^2)/9-((y-2)^2)/(22)=1

foci

\frac{( x - 3 ) ^{2}}{9} - \frac{( y - 2 ) ^{2}}{22} = 1

(3 + 31, 2), (3 - 31, 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{9} - \frac{( y - 2 ) ^{2}}{22} = 1 :

Right-left Hyperbola with

(h, k) = (3, 2), a = 3, b = 22

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

Compute

c :

c = 3^{2} + 22^{2} : 31

(3 + 31, 2), (3 - 31, 2)

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

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

\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

What is the foci ((x-3)^2)/9-((y-2)^2)/(22)=1 ?
The foci ((x-3)^2)/9-((y-2)^2)/(22)=1 is (3+sqrt(31),2),(3-sqrt(31),2)