What is the vertices (x^2)/(16)-(y^2)/(48)=1 ?

The vertices (x^2)/(16)-(y^2)/(48)=1 is (4,0),(-4,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

vertices (x^2)/(16)-(y^2)/(48)=1

vertices

(4, 0), (- 4, 0)

Solution steps

(h + a, k), (h - a, k)

Calculate Hyperbola properties

\frac{x ^{2}}{16} - \frac{y ^{2}}{48} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 43

(0 + 4, 0), (0 - 4, 0)

Refine

(4, 0), (- 4, 0)

What is the vertices (x^2)/(16)-(y^2)/(48)=1 ?
The vertices (x^2)/(16)-(y^2)/(48)=1 is (4,0),(-4,0)