How to determine b in a conic hyperboal graph

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Homework Help Overview

The discussion revolves around determining the value of "b" in the equation of a hyperbola, specifically in the context of a conic section graph. The original poster expresses difficulty in finding a method to solve for "b" without a specific point on the graph.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the necessity of having a point off the x-axis to solve for "b". One suggestion involves using the coordinates of a point near the graph to approximate "b". Another participant introduces the concept of asymptotes and their relationship to finding "b", indicating that different points may yield varying results.

Discussion Status

The discussion is active, with participants exploring different methods to determine "b". There is acknowledgment that using asymptotes may provide an alternative approach, although participants are relying on visual estimation from the graph.

Contextual Notes

Participants note the challenge of solving for "b" without specific points and the reliance on graphical interpretation, which may lead to different approximations. The original poster's request for a solution without a point highlights the constraints of the problem.

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How to determine "b" in a conic hyperboal graph

Homework Statement



http://img143.imageshack.us/img143/3391/91667159.jpg
(x-1)2/22 - y2/b2 =1
I can't find any good point for me to solve b..I don't know what to do..
Is there any way to solve b without using the point?

Homework Equations


The Attempt at a Solution

 
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Not really. You need a point off the x-axis to solve for b. It looks like it passes near the point (4,1). That will let you get an approximate value for b.
 


I see, thanks!
 


What you really need, in order to find b, is the equation of asymptotes. If a hyperbola has equation
\frac{(x-x_0)^2}{a^2} - \frac{(y-y_0)^}{b^}= 1
then its asymptotes are y-y_0= \pm b(x-x_0)/a

On this graph, it looks to me like an asymptote passes through (1,0) and (3,1) so has equation y= (1/2)(x- 1). That gives you a slightly different answer than assuming the graph passes through (4,1) but Dick and I are both "eyeballing" the graph.
 

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