# How do you find the center, vertices, foci and asymptote of #2x^2-3y^2=6#?

##### 1 Answer

**Centre** **Vertices are**

**Focii are** **Asymptote is**

#### Explanation:

This is standard form of the equation of a hyperbola with center

The vertices are

Vertices are

Each hyperbola has two asymptotes that intersect at the center of

the hyperbola. The asymptotes pass through the vertices of a

rectangle of dimensions

centre at

asymptote is

graph{2x^2-3y^2=6 [-10, 10, -5, 5]} [Ans]