Intersection of Line and Hyperbola

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September 29th 2011, 01:13 AM
moriman

Intersection of Line and Hyperbola

A hyperbola is intersected by line 1 at $(u_1, v_1)$ and $(v_1, u_1)$
and by line 2 at $(u_2, v_2)$ and $(-v_2, -u_2)$
Lines 1 and 2 intersect at $(u_1+v_1, 0)$

These can be written

$u_1+v_1=u_2-v_2\ \ \ \ \ \ \(1)$
$u_1v_1=u_2v_2\ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

I am trying to find a general solution for the values of $u_1, v_1, u_2$ and $v_2$ .

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
September 29th 2011, 04:19 AM
moriman

re: Intersection of Line and Hyperbola

ok,

Here's where I am so far...

$u_1+v_1=u_2-v_2$

$u_1v_1=u_2v_2$

$let\ \ \ u_2=u_1+x$

$and\ \ v_2=v_1-y$

then we need

$(u_1+x)(v_1-y)=u_1v_1$

$v_1x-xy-u_1y=0$

$y=\dfrac{v_1x}{u_1+x}$

$x=\dfrac{u_1y}{v_1-y}$

Is this as far as I can go?
September 29th 2011, 10:36 AM
moriman

Re: Intersection of Line and Hyperbola

I had forgotten about one of the original constraints

$u_1+v_1=u_2-v_2$

This works out to $x=2v_1-y$

Taking this with the $x$ from the previous result I have

$2v_1-y=\dfrac{u_1y}{v_1-y}$

This gives

$y=\dfrac{(3v_1+u_1)\pm\sqrt{u_1^2+6u_1v_1+v_1^2}}{ 2}$

So for integer solutions $(u_1^2+6u_1v_1+v_1^2)$ must be a perfect square.
September 29th 2011, 11:19 AM
Siron

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? ...
September 29th 2011, 11:32 AM
moriman

Re: Intersection of Line and Hyperbola

Hi,

With the original equations

$u_1+v_1=u_2-v_2$

$u_1v_1=u_2v_2$

The only way to find values for $u_1, v_1, u_2, v_2$ 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

$y=\dfrac{(3v_1+u_1)\pm\sqrt{u_1^2+6u_1v_1+v_1^2}}{ 2}$

and whatever $x$ 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
September 30th 2011, 12:51 AM
moriman

Re: Intersection of Line and Hyperbola

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

Take $a^2+b^2=c^2$

If one sets

$a=m^2-1$ and $b=2m$ then $c=m^2+1$

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.

$u_1+v_1=u_2-v_2$

$u_1v_1=u_2v_2$

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

u_2 is not only greater than u_1, but also $u_2>(u_1+v_1)$

$so I set\ \ \ \ \ u_2=(u_1+v_1)+x$

$and\ \ \ \ \ \ \ \ \ \ v_2=x$

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

$u_1+v_1=\dfrac{m-n}{2}$

$u_1v_1=\dfrac{mn}{4}$

$and\ \ \ x=\dfrac{n}{2}$

which is no further forward :(

Quick synopsis of where I am at the moment

$u_1+v_1=u_2-v_2$

$u_1v_1=u_2v_2$

$u_2-(u_1+v_1)=\dfrac{u_1v_1}{u_2}$

$u_2^2-(u_1+v_1)u_2-u_1v_1=0$

$u_2=\dfrac{(u_1+v_1)\pm\sqrt{(u_1+v_1)^2+4u_1v_1}} {2}$

similarly

$v_2=\dfrac{-(u_1+v_1)\pm\sqrt{(u_1+v_1)^2+4u_1v_1}}{2}$

Solving for $(u_1+v_1)^2+4u_1v_1$ being a perfect square leads me back to the beginning.