Intersection of Line and Hyperbola

**moriman** · September 29th 2011, 01:13 AM

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

**moriman** · September 29th 2011, 04:19 AM

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?

**moriman** · September 29th 2011, 10:36 AM

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.

**Siron** · September 29th 2011, 11:19 AM

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

**moriman** · September 29th 2011, 11:32 AM

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

**moriman** · September 30th 2011, 12:51 AM

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

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.

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