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Lines Tangent to a Hyperbola

Date: 07/14/98 at 02:57:05
From: Bryan J. 
Subject: Tangent lines to graphs

Hi again Dr. Math,

Here's a problem I can't exactly figure out.

   x^2 - y^2 = 1
           y = 2x + k

I did substitution and all that and got:
  3x^2 + 4kx + (k^2 +1) = 0

But I can't figure out where to go from here. Could you give me a 
hand or hint? I can't figure out how to sift the k from the x and y. 
It may be math I'm rusty on or something, but it escapes me at the 
moment. Thanks a bunch.


Date: 07/22/98 at 11:19:51
From: Doctor Rick
Subject: Re: Tangent lines to graphs

Hi, Bryan. 

You've done well so far. Now maybe you need to back off and remind 
yourself where you're going. You've got two equations in three 
unknowns, reduced to one equation in two unknown. You can't solve it 
for all the unknowns - but from your subject line, I get some 
additional information.

Did you sketch out the equations first? This can help a lot when you 
don't know where you're going. You've got a hyperbola, and a line of 
fixed slope with the y-intercept as a parameter. I take it that the 
problem is to find the value(s) of k such that the line is tangent to 
the hyperbola.

        o                    |      /       /     o      
          o                  |     /       /    o      
            o                |    /       /   o         
              o              |   /       /  o           
                o            |  /       / o        
                  o          | /       /o              
                    o        |/       o                
                     o       + k     o               
                      o     /|      o                  
                       o   / |     o                   
                       o  /  |    /o                    
                       o/    |  /  o                    
                       o     | /   o                    
                      o      |/     o                  
                     o       + -k    o               
                    o       /|        o                
                  o/       / |          o              
                o /       /  |            o        
              o  /       /   |              o           
            o   /       /    |                o         
          o    /       /     |                  o      
        o     /       /      |                    o     
Now, when the line is tangent to the curve, what will be true? There 
will be only one point of intersection between the line and the 
hyperbola! At all other points, there will be either two solutions, or 
no (real) solutions.

You have derived a quadratic equation in x, with k as a parameter. 
Solve for x using the quadratic formula. What has to happen for there 
to be a single solution? This will give you an equation in k alone; 
solve it and you are done. (Note from my sketch that there should be 
two solutions. Remember to check your solutions by plugging them back 
in the original equations.)

If this help doesn't get you through the problem, write again.

- Doctor Rick, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   

Date: 07/15/98 at 18:07:12
From: Bryan J. 
Subject: Correction to a problem on tangents

Actually, in my haste, I forgot to include part of the problem. The 
original one was if y = 2x + k is a line tangent to the graph 
x^2 + y^2 = 1, what is k? Then I did all the substitution I showed 
above. Well anyway, I figured out the problem. If there's only one 
solution (tangency), then b^2  - 4ac = 0. So:

   (4k)^2 - 4(3)(k^2 + 1) = 0 
   4k^2 = 12
   k = +sqrt(3) or -sqrt(3)
Associated Topics:
High School Equations, Graphs, Translations

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