Asymptotes of Hyperbolas (1 Viewer)

[nrumble42]

(x-3)^2/64- (y+1)^2/36=1

How would you find the asymptotes of this? I can't seem to get it right.

Thanks guys!

 

[calamebe]

Firstly, just consider the asymptotes of the hyperbola x^2/64-y^2/36=1. This would just be y = ±3x/4. Now, to find the asymptotes of (x-3)^2/64 - (y+1)^2/36 = 1, notice that it is just the hyperbola x^2/64-y^2/36=1 but translated 3 to the right and 1 down. So just replace x with x-3 and y with y+1 to get the asymptotes as y+1=±3(x-3)/4 or y = 3x/4 - 13/4 and y = -3x/4 + 5/4.

 

[InteGrand]

Also to find asymptotes for hyperbolas "centred" at the origin: just set the RHS of the hyperbola equation to 0 and rearrange to get the asymptotes' equations. (This technique also works if the hyperbola is of the form y^2/(A^2) - x^2/(B^2) = 1 so you don't need to learn different formulas for the asymptotes of this type of hyperbola.)