2
$\begingroup$

How can we parametrize the conic $C$: $x^2+y^2 = 5$, by considering a variable line through $(2,1)$ and hence all rational solutions of $x^2 + y^2 = 5$?

I'm thinking let $x = \sqrt{5}\cos t$, and $y = \sqrt{5} \sin t$, and somehow I need to get the coordinates $(2,1)$ in also.

$\endgroup$
  • 2
    $\begingroup$ This is not a precalculus question $\endgroup$ – mary Aug 14 '12 at 19:59
8
$\begingroup$

What you need to do is this:

Find the equation of a line through $(2,1)$ whose gradient $t$ is a rational number. You should get $y-1 = t(x-2)$.

Take a moment to visualise what is happening. As $t$ increases from large and negative, to large and positive, this line rotates about $(2, 1)$, through almost a half turn. We need to find the other intersection of the line and the conic.

We can do that by substituting for $y$, in your original equation; that gives us a quadratic equation in $x$, for any given value of $t$. We already know one of the roots of this quadratic, since it must be satisfied by $x=2$, since the line and your conic both go through $(2,1)$.

The quadratic you get is $$x^2(1+t^2) + x(-4t^2+2t) + 4t^2-4t-4 = 0$$

Using algebraic long division, we can therefore factorise this to get $$\big[x-2\big]\big[x(1+t^2) -2t^2+2t+2\big] = 0$$

So the other root is given by $$x = \frac{2(t^2-t-1)}{1+t^2}$$ and then, using the fact that $y = tx-2t+1$ with a bit of algebra, we get

$$y = \frac{-t^2-4t+1}{1+t^2}$$

You can double-check that $x^2+y^2=5$, and thus we have a parametrisation giving the rational points on your curve, by taking rational values for $t$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm not sure about the "gradient", do you mean slope? $\endgroup$ – mary Aug 14 '12 at 20:19
  • $\begingroup$ Yes, gradient = slope. $\endgroup$ – Old John Aug 14 '12 at 20:20
  • 1
    $\begingroup$ how did you get to the x^2(1+t^2) + x(-4t^2+2t) portion, I wasn't able to get to there/understand it from your work. Thx $\endgroup$ – mary Aug 14 '12 at 20:47
  • 1
    $\begingroup$ What I did was to substitute $y = tx-2t+1$ into the equation $x^2 + y^2 = 5$, and rearranged it a bit. $\endgroup$ – Old John Aug 14 '12 at 20:49
  • 2
    $\begingroup$ One can bypass factoring by noting that the sum of the roots is $\frac{4t^2-2t}{1+t^2}$, and subtracting $2$. $\endgroup$ – André Nicolas Aug 14 '12 at 21:16
2
$\begingroup$

In General formula generic for Pythagorean triples looks a little different.

$$x^2+y^2=az^2$$

If the number can be represented as a sum of squares. $a=t^2+k^2$

The solution has the form:

$$x=-tp^2+2kps+ts^2$$

$$y=kp^2+2tps-ks^2$$

$$z=p^2+s^2$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.