Hyperbola and lines parallel to its asymptotes

[Shadow]

Let h be a hyperbole given by the equation:

 

[math]h : \frac{(x-m)^2}{a^2} - \frac{(y-n)^2}{b^2} = 1[/math]

 

Its asymptotes are then:

 

[math] a_{1, 2}: y-n= \pm \frac{b}{a} (x-m)[/math]

 

We were told in our math class that every line parallel to the asymptotes of a hyperbola intersects the hyperbola in only one point. So let us define a line l which is parallel to one of the asymptotes:

 

[math]l: y=\frac{b}{a}x + c[/math]

 

From what I've been told, this line should intersect the hyperbola only once for every real value of c excluding the ones for which [math] l = a_{1, 2}[/math]. Let's pretend we don't know that, and try to calculate c so that the resulting line intersects the hyperbola in only one point.

 

I'm not sure if this is the method that's generally used to calculate c, but this is what they taught us. If we substitute y in the equation of h for [math]\frac{b}{a}x + c[/math]:

 

[math]\frac{(x-m)^2}{a^2} - \frac{(\frac{b}{a}x + c-n)^2}{b^2} - 1 = 0[/math]

 

we are effectively finding the intersection points of h and l. If we expand the left side of the above equation, we (should) get a parametrized quadratic equation with the parameter c. Since we want the line to only have one intersection point with the hyperbola, we want the parameter to be such that there will only be one (multiple) solution, ie. the discriminant of this quadratic polynomial must be zero. So we take the discriminant D with respect to x, and solve the equation

 

[math]D=0[/math]

 

Now, in accordance with the fact that every line parallel to one of the asymptotes only has one intersection point with the hyperbola, the above equation should have an infinite number of solutions, excluding the one for which the line is the asymptote. Unfortunately, what happens is that the terms with [math]x^2[/math] in them cancel out, and what should have been a parametrized quadratic equation becomes a parametrized linear equation. The discriminant of a linear polynomial is equal to 1, and this is my first question: why is the discriminant of a linear equation equal to one?

 

Regardless, the equation [math]1=0[/math] has no solutions, which basically means that there exists no c for which the line [math]y=\frac{b}{a}x + c[/math] intersects the hyperbola only once. So my question is, why doesn't this method of calculating c work when dealing with lines parallel to the asymptotes?

 

Thanks

Edited August 6, 2010 by Shadow

[DJBruce]

I might be mistaken but I think you made a mistake when you define a line parallel to one of the asymptotes as:

 

[math] l: y=\frac{b}{a}+c[/math]

 

this would simply be a horizontal line. I believe you are wanting

 

[math] l: y=\frac{b}{a}x+c[/math]

 

which would mean that your substitution would then be incorrect.

[Shadow]

Sorry, that's just a typo, the calculations were performed using the correct equation for a line.

 

Corrected.

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Merged post follows:

Consecutive posts merged

Sorry for the bump, but this has been bugging me for a while and I'd be extremely grateful for any help.

Edited June 14, 2010 by Shadow