# Asymptotes for hyperbolas q2 problem :S

1. Aug 12, 2006

### Chadlee88

Does the equation (x+1)^2-4y^2 = 0 have asymptotote??
i graphed it and from the graph it does not look like a hyperbola because it seems to intersect at the point x = -1 y = 0

Thanks HallsofIvy for helpin me with the previous asymptote hyperbola prob

2. Aug 13, 2006

### WigneRacah

The equation you wrote doesn't describe a hyperbola but two straight lines intersecting in the point (-1, 0). In fact you can write

$$(x+1)^2- 4 y^2 = 0 \Rightarrow (x +1 - 2 y) (x +1 + 2 y) =0$$

which implies

$$x+1 -2y=0$$

or

$$x + 1 + 2 y =0$$

(the stright line equations).

If the equation was

$$(x+1)^2-4y^2=1$$

it described a hyperbola shifted back along the x-axis with vertex in (-1,0).
The previously stright line equations are the asymptotes.

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