Intersection of Line and Hyperbola

A hyperbola is intersected by line 1 at and

and by line 2 at and

Lines 1 and 2 intersect at

These can be written

I am trying to find a general solution for the values of and .

I have tried squaring (1) and adding\subtracting various multiples of (2), but this just brings me various fractions of the starting values.

Any help or pointers would be greatly appreciated

re: Intersection of Line and Hyperbola

ok,

Here's where I am so far...

then we need

Is this as far as I can go?

Re: Intersection of Line and Hyperbola

I had forgotten about one of the original constraints

This works out to

Taking this with the from the previous result I have

This gives

So for integer solutions must be a perfect square.

Re: Intersection of Line and Hyperbola

I have red your posts and it's confusing in my opinion, what do you want to calculate or show? ...

Re: Intersection of Line and Hyperbola

Hi,

With the original equations

The only way to find values for is by trial and error.

I am trying to find a general solution.

It is similar to the diophantine solution for a pythagorean triple

In the attempt I've made so far, the

and whatever will work out to be ( a similar expression to y)

If I can make sure that x and y are integers, I would have correct numbers to satisfy the original equations

Re: Intersection of Line and Hyperbola

OK, I'll try giving an example of what I'm looking for

Take

If one sets

and then

By choosing *any* integer m>1 a, b and c will be a pythagorean triple.

This doesn't give *all* pythaorean triples, but it does give an infinite amount of them, which is usually enough :D

I'm looking for a similar way of getting numbers that satisfy the 2 equations in the op.

I had made further progress from my last post after I noticed that

u_2 is not only greater than u_1, but also

This satisfies the first equation. When solving with the second equation I would have

which is no further forward :(

Quick synopsis of where I am at the moment

similarly

Solving for being a perfect square leads me back to the beginning.