What is the foci ((x+1)^2)/9-((y-3)^2)/(16)=1 ?

The foci ((x+1)^2)/9-((y-3)^2)/(16)=1 is (4,3),(-6,3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x+1)^2)/9-((y-3)^2)/(16)=1

foci

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{16} = 1

(4, 3), (- 6, 3)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (- 1, 3), a = 3, b = 4

(- 1 + c, 3), (- 1 - c, 3)

Compute

c :

c = 3^{2} + 4^{2} : 5

(- 1 + 5, 3), (- 1 - 5, 3)

Refine

(4, 3), (- 6, 3)

\frac{x ^{2}}{25} - \frac{y ^{2}}{64} = 1

\frac{x ^{2}}{4} - \frac{y ^{2}}{12} = 1

x^{2} - y^{2} = 9

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

\frac{x ^{2}}{16} - \frac{y ^{2}}{64} = 1

What is the foci ((x+1)^2)/9-((y-3)^2)/(16)=1 ?
The foci ((x+1)^2)/9-((y-3)^2)/(16)=1 is (4,3),(-6,3)