Math Is Fun Forum

1. Find the angle between the asymptotes to the hyperbola 3x² - 5xy -2y² + 17x + y + 14 = 0.

This is honestly the most lengthy, involved problem I've ever worked on.  I never studied the general conic equation while in school, so this is new for me.  Hopefully I've done everything correctly.
First we need to get rid of the xy term by doing a rotation of axes..

using the formulae for the axes of rotation

We then substitute these into the given equation and after some work we get..

The x'y' terms cancel and after much, much more work we end up with...
(I changed x' and y' to simply x and y)

Now here I had a little help from the computer and amazingly...

continuing further along we eventually end up with...

So the slope of the asymptotes are

The hyperbola is oriented parallel to the y-axis and to get the angle between the asymptote and x-axis we use

So the angle between the asymptotes is

or on the other side the angle is

I hope I did everything right.  I wouldn't wish this problem upon my worst enemy.

Welcome aboard Aesdyn2!
      I agree there are some brilliant and knowledgeable people on this site.  This is great site for learning and practicing problems.   I'm not in school so I try to learn what I can.
       Fruityloop

Good job Bobbym!

Oops...boy was I wrong.
(a+b)(a^2-ab+b^2)=3(2007-ab) doesn't mean (a+b)=3 or (a^2-ab+b^2)=3

Very good!
The method I used was unnecessarily complicated.
I overlooked using X=2^x in the quadratic equation.

The graph is simply showing all values of X that satisfy both X<-2 and X<3.  The open dot simply means that -2 isn't part of the solution, but everything less than -2 is part of the solution.

From the first sentence one can get..

from the second sentence one can get..

where E and H are Edna's and Henry's current ages respectively.
solve for one and substitute in the other for the answer.
Thanks phrontister! Good confusing problem.

This post is a few years old, but I find this problem fascinating...
It reminds me a little bit of the game of c r a p s where you have to roll your 'point' before a 7 in order to win.  You have a infinite number of possible sequence of events depending on when the 'point' or 7 appears.  Using a nice diagram like Jane's...

Ben wins with the far branch on the right H-T-H or with the branch ending on the bottom left H-H-T-H.
Bill wins with the branch ending on the middle left H-H-H.
So the probability of Ben winning is (1/8+1/16)=3/16 and Bill's chances are (1/8)=2/16.
So, Ben's chances are 3/16 and Bill's chances are 2/16.
So, Ben is favored 3 to 2 with a fair coin.  Ben has a 60% chance of winning and Bill has only a 40% chance of winning.
We want to find the probability of heads which will give an even chance to both parties.
Let the probability of heads be x and the probability of tails be 1-x.
Bill must win with three heads and Ben can win with either two heads and a tail or three heads and a tail so...

which gives the solution as

or about 61.8% chance of the coin being heads.

Hi Fla$h!;
Here's #2...

using

we get

I am not 100% sure of myself but after a rather lengthy process I got...

I used volume by slicing using

as the radius of the disks and then subtracting out the middle cylinder which has a radius of k and a height of 2r.

We've just had the hottest day ever here in the Pacific Northwest of the U.S. on July 29.  The city I live in reached 104°F or 40°C!!!!

Somebody on T.V. said this isn't because of global warming!?!?!?

I realize it's only one day, but the temperatures are off the chart of what's normal.

Bobbym,
     How in the world did you figure that out?  Somehow, you knew to choose 3 and 4 and then figured out the irrational root.
Highly impressive.
     Thank you for your time.

Bobbym's answer is given as being correct on the website but it doesn't seem correct.
I guess the reasoning is that if we have five numbers and if we cycle through all combinations for three of them and multiply that by the different combinations of three out of five we will have the answer.   So we have 1000(000-999)*5C3 = 10,000.
So 123XX will be different from XX311, but we can choose 11 for the first number and 12 for the second number and we have the same number!
Each number selected eliminates 45 other numbers, so the total number of possible IDs is only 1/46 of the total. So 100000/46 = 2173 possible IDs.  I'm not totally sure my answer is correct.
Anybody else have any ideas?