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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.

Bryan
```

```
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

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      |/     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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