What is the foci ((x+2)^2)/9-((y-1)^2)/4 =1 ?

The foci ((x+2)^2)/9-((y-1)^2)/4 =1 is (-2+sqrt(13),1),(-2-sqrt(13),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)/4 =1

foci

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

(- 2 + 13, 1), (- 2 - 13, 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

c = 3^{2} + 2^{2} : 13

(- 2 + 13, 1), (- 2 - 13, 1)

x^{2} - y^{2} = 13

\frac{( y ) ^{2}}{81} - \frac{( x ) ^{2}}{121} = 1

x^{2} - y^{2} = 15

x^{2} - 10 x = 4 y^{2} - 5

2 x^{2} - 2 y^{2} - 20 x + 16 y - 16 = 0

What is the foci ((x+2)^2)/9-((y-1)^2)/4 =1 ?
The foci ((x+2)^2)/9-((y-1)^2)/4 =1 is (-2+sqrt(13),1),(-2-sqrt(13),1)