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For a National Board Exam Review:

Find the Equation for the Asymptotes of a Hyperbola ${ (y-x)^2 - (x+5)^2 = 36 }$

Answer is ${ y-5 = \pm (x+5) }$

I've already solved the equations: here they are:

$${ y = x+10 }$$ $${ y = -x }$$

My problem is how to translate it into this " ${ y-5 = \pm (x+5) }$ " ? I know that if you reverse engineer the equation by doing seperate equations for each you could end up with my answer... But I want to know if there is a method for methodically translating my answer to the one with the ${\pm}$ sign on it...

james
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3 Answers3

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You already got the geometric answer, here is the algebraic one:

You have the equations:

$$y=ax + b\\ y=-ax - c$$

You can rearange the second one into:

$$y=-(ax + \frac{b+c}{2} + c - \frac{b+c}{2})$$

Which rearanges into $$y-\frac{b-c}{2} = -(ax+\frac{b+c}{2})$$

Similarly, the second equation rearanges from

$$y=ax + \frac{b+c}{2} + b - \frac{b+c}{2}$$

Into $$y-\frac{b-c}{2} = ax + \frac{b+c}{2}$$

5xum
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Just draw the lines in a coordinate system; they will intersect at $(x,y)=(-5,5)$, so in terms of the translated coordinate system centered at that point, $(u,v)=(x+5,y-5)$, their equations will be $u = \pm v$.

Hans Lundmark
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Using $$b=\frac{b+c}{2}+\frac{b-c}{2}\quad\text{and}\quad c=\frac{b+c}{2}-\frac{b-c}{2},$$ we can see that $$y=ax+b$$ $$y=-ax+c$$ can be written as $$y=ax+\frac{b+c}{2}+\frac{b-c}{2}\Rightarrow y-\frac{b+c}{2}=ax+\frac{b-c}{2}$$ $$y=-ax+\frac{b+c}{2}-\frac{b-c}{2}\Rightarrow y-\frac{b+c}{2}=-ax-\frac{b-c}{2},$$ i.e. $$y-\frac{b+c}{2}=\pm\left(ax+\frac{b-c}{2}\right)$$

Your case is $(a,b,c)=(1,10,0)$.

mathlove
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