What is the foci ((x+2)^2}{16}-\frac{(y-1)^2)/9 =1 ?

The foci ((x+2)^2}{16}-\frac{(y-1)^2)/9 =1 is (3,1),(-7,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

foci

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

(3, 1), (- 7, 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (- 2, 1), a = 4, b = 3

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

Compute

c :

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

(- 2 + 5, 1), (- 2 - 5, 1)

Refine

(3, 1), (- 7, 1)

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

\frac{x ^{2}}{36} - \frac{y ^{2}}{25} = 1

y^{2} - 4 x^{2} = 36

\frac{( y - 2 ) ^{2}}{14} - \frac{( x + 6 ) ^{2}}{3} = 4

3 x^{2} - 13 y^{2} - 30 x + 52 y = 16

What is the foci ((x+2)^2}{16}-\frac{(y-1)^2)/9 =1 ?
The foci ((x+2)^2}{16}-\frac{(y-1)^2)/9 =1 is (3,1),(-7,1)