What is the foci ((x-6)^2}{16}-\frac{(y-2)^2)/9 =1 ?

The foci ((x-6)^2}{16}-\frac{(y-2)^2)/9 =1 is (11,2),(1,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

foci

(11, 2), (1, 2)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (6, 2), a = 4, b = 3

(6 + c, 2), (6 - c, 2)

Compute

c :

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

(6 + 5, 2), (6 - 5, 2)

Refine

(11, 2), (1, 2)

What is the foci ((x-6)^2}{16}-\frac{(y-2)^2)/9 =1 ?
The foci ((x-6)^2}{16}-\frac{(y-2)^2)/9 =1 is (11,2),(1,2)