What is the foci ((x-2)^2)/9-((y+1)^2)/5 =1 ?

The foci ((x-2)^2)/9-((y+1)^2)/5 =1 is (2+sqrt(14),-1),(2-sqrt(14),-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x-2)^2)/9-((y+1)^2)/5 =1

foci

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

(2 + 14, - 1), (2 - 14, - 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (2, - 1), a = 3, b = 5

(2 + c, - 1), (2 - c, - 1)

Compute

c :

c = 3^{2} + 5^{2} : 14

(2 + 14, - 1), (2 - 14, - 1)

x^{2} - y^{2} \geq 1

\frac{( x - 6 ) ^{2}}{16} - \frac{( y - 2 ) ^{2}}{9} = 1

\frac{y ^{2}}{16} - \frac{x ^{2}}{25} = 1

- x^{2} + 4 y^{2} - 2 x - 16 y + 11 = 0

9 x^{2} - 16 y^{2} - 18 x - 64 y - 199 = 0

What is the foci ((x-2)^2)/9-((y+1)^2)/5 =1 ?
The foci ((x-2)^2)/9-((y+1)^2)/5 =1 is (2+sqrt(14),-1),(2-sqrt(14),-1)