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 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 |/ 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)
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