Question

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

Answer 1

Centre `(0,0)`, Vertices are `(-sqrt3,0) and (sqrt3,0)`,
Focii are `(-sqrt5,0) and (sqrt5,0)` Asymptote is
`y= +-sqrt(2/3)x`

Explanation:

`2x^2-3y^2=6 or (2x^2)/6-(3y^2)/6=1` or

`x^2/3-y^2/2=1 or (x-0)^2/(sqrt3)^2 -(y-0)^2/(sqrt2)^2=1`

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

`(h,k) ; (x-h)^2/a^2-(y-k)^2/b^2=1 :` Centre `(0,0)`

The vertices are `a` units from the center, and the foci are

`c` units from the center. Moreover `c^2=a^2+b^2`

Vertices are `(-a,0) and (a,0) ; a = sqrt3 and b=sqrt2` or

`(-sqrt3,0) and (sqrt3,0) ; c^2= (sqrt3)^2+sqrt(2)^2=5`

`:. c= +- sqrt5 :.` Focii are at `(-sqrt5,0) and (sqrt5,0)`

Each hyperbola has two asymptotes that intersect at the center of

the hyperbola. The asymptotes pass through the vertices of a

rectangle of dimensions `2a=2sqrt3 and 2b=2sqrt2` with its

centre at `(0,0) :.` slope `=+- (sqrt2/sqrt3)` Equation of

asymptote is `y-0= +-sqrt(2/3)(x-0) or y= +-sqrt(2/3)x`

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