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Question 27826: My question is for hyperbolas,
I need to fine the vertecies foci slope and then draw a graph, the problem is i was sick the day my Algebra teachr went through it and my personal oppinion he should not be teaching math because he doesnt answere my questions! one of the many equations is: (x+6) quantity squared over 25, minus (y+3) quantity squared over 9 quals(=) 1(one) please help me get this stuff!
-Sick and Desperate
Answer by bmauger(101)   (Show Source):
You can put this solution on YOUR website! 
So the major axis (positive value) is x. a = root of 25 = 5, b = root of 9 = 3
 approximately 5.830
foci = (k plus/minus c, h) = -6+5.830= (-0.169, -3) and -6-5.830 = (-11.830, -3)
for slope you want b/a and -b/a when x is your major axis or a/b and -a/b when y is your major axis (in other words, whichever value is under the y^2 term should be on top, since (change in y/change in x) is slope. For your problem, the slope will be 5/3 and -5/3. The lines of the asymptotes have the slope 3/5 or -3/5 and go through the point (k, h). The equation for that line is:
For your problem:
 and 
 and 
To graph the equation first plot the verticies. Those are the a values (k+a, h) and (k-a, h). This is -6+5=(-1, -3) and -6+5=(-11,-3) Next graph the asymptotes. Then draw a hyperbola that starts at the verticies and fits between the asymptotes approaching them, but not crossing as you go out. (usually teachers just want it to look pretty good and for you to apply the right methods.) Kinda like this:
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