# how do i find the center, transverse axis, vertices, foci and asymptotes for the hyperbola? 2x^2-y^2=4help

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### 1 Answer

Given the hyperbola `2x^2-y^2=4` :

The general form is `(x-h)^2/a^2-(y-k)^2/b^2=1` for a hyperbola with a horizontal major axis. The center will be at (h,k).

For hyperbolas centered at the origin, the asymptotes are given by `y=+-b/ax` . The transverse axis will have length 2a. The vertices will be at (-a,0),(a,0). The foci will be at (-c,0),(c,0) where `c^2=a^2+b^2` .

Putting the given equation in general form we have:

`x^2/2-y^2/4=1` So `a=sqrt(2),b=2`

(1) The center is at (h,k) which is (0,0) or the origin for this hyperbola.

(2) The transverse axis is 2a or `2sqrt(2)`

(3) The vertices are at (-a,0),(a,0) or `(-sqrt(2),0),(sqrt(2),0)`

(4) To find the foci we solve `c^2=a^2+b^2;c^2=2+4==>c=+-sqrt(6)` so the foci are at `(-sqrt(6),0),(sqrt(6),0)`

(5) The asymptotes are `y=+-2/sqrt(2)x`

Solving for y we get `y=+-sqrt(2x^2-4)` ; the graph: