So I have the equation defined in homogeneous coordinates $[w; x, y]$ as $[1+t^2; 1-t^2, 2t]$

$$w = 1+t^2$$ $$x = 1-t^2$$ $$y = 2t$$

If I do $w+x-y$ I get $-2t+2$, so $t = -(w+x-y-2)/2$. I was then going to plug this back into one of the original parameters and set equal to $0$.

So $2t = 0 $

$${-(w+x-y-2)}/{2} = 0$$ $$-(w+x-y-2) = 0 $$ $$-w + x - y +2 = 0$$ $$ x - y - w +2 = 0$$

And that would be my final equation.

However I feel like this isn't right because I feel like I shouldn't have the 'w' term in there at all. Originally I was going to change back into Cartesian Coordinates and divide $x$ and $y$ by $w$ but I found the new equations to be more cumbersome so I decided to try it this way.

So am I doing something wrong or is this correct?

  • $\begingroup$ At the top of your question you have $[w,x,y]=[1+t^2,1-t^2,2t].$ But then in your displayed equations after that you have the formula $y=1+2t$. But one of these is wrong, so which is it? That is, at the top of the question, $y=2t$ not $y=2t+1$. Also with your $w,x,y$ from the rest of the question, the value of $w+x-y$ is $-2t+1$ and not $-2t-1$ as you have it. $\endgroup$ – coffeemath Feb 1 '13 at 23:59
  • $\begingroup$ Oh nice catch, I copied down the equation wrong. I'll fix it. However is my approach correct? Is this generally how you'd go about solving this? Or should I convert it back into cartesian coordinates first so I don't have a 'w' term? $\endgroup$ – user1855952 Feb 2 '13 at 0:10
  • $\begingroup$ With the updated definitions, you are correct at $w+x-y=-2t+2$. However the algebra after that must be off, because actually $x-y-w+2=-2t^2+2t+2$. I suspect there cannot be a linear equation satisfied by $w,x,y$ of any coefficients, that is $aw+bx+cy=(a-b)t^2+2ct+a+b$ and for this to be $0$ we get $a=b=c=0$ and the equation is $0w+0x+0y=0$, which cannot be regarded as the right relation, since it is "trivial" and in fact holds for all choices of $x,y,z$. $\endgroup$ – coffeemath Feb 2 '13 at 1:12

If the definition is as at the top of the question $[w,x,y]=[1+t^2,1-t^2,2t]$ then you can compute that $$x^2+y^2-w^2=0$$ by plugging in for $w,x,y$ their parametric expressions.

If the definition is really $[w,x,y]=1+t^2,1-t^2,2t+1]$ then the above equation doesn't work, because $x^2+y^2-w^2=4t+1$ in that case, and the parameter $t$ has not been eliminated.

I would guess it should be as at the top of the question with $2t$ for $y$, as then this is a standard way to give a parametric expression for $x^2+y^2=w^2$.

  • $\begingroup$ How were you able to arrive at x^2 + y^2 - w^2? I understand why it's correct because I plugged in the values to check but how did you intuitively or methodologically arrive at that answer? $\endgroup$ – user1855952 Feb 2 '13 at 0:36
  • $\begingroup$ The three functions of $t$ are fairly often used to parametrize the circle $x^2+y^2=1$. For this the equations are obtained by dividing through by $w=1+t^2$ to get $x=(1-t^2)/(1+t^2)$ and $y=2t/(1+t^2).$ With these as $x,y$ the same check shows that $x^2+y^2=1$. $\endgroup$ – coffeemath Feb 2 '13 at 0:59

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