What is the vertices (x^2)/2-(y^2)/3 =1 ?

The vertices (x^2)/2-(y^2)/3 =1 is (sqrt(2),0),(-sqrt(2),0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (x^2)/2-(y^2)/3 =1

vertices

\frac{x ^{2}}{2} - \frac{y ^{2}}{3} = 1

(2, 0), (- 2, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{2} - \frac{y ^{2}}{3} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 2, b = 3

(0 + 2, 0), (0 - 2, 0)

Refine

(2, 0), (- 2, 0)

- 4 x^{2} + y^{2} + 16 x + 12 y + 16 = 0

\frac{y ^{2}}{81} - \frac{x ^{2}}{64} = 1

\frac{( x ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{9} = 1

4 (y - 3)^{2} - 9 (x - 2)^{2} = 36

\frac{x ^{2}}{36} - \frac{( y - 10 ) ^{2}}{64} = 1

What is the vertices (x^2)/2-(y^2)/3 =1 ?
The vertices (x^2)/2-(y^2)/3 =1 is (sqrt(2),0),(-sqrt(2),0)