Question 3702
i cannot eliminate t from these, but i can get t=, so then you can put this back into your equations...


{{{yb = (2t)/(1-t^2)}}}
{{{yb = (t^2+2t+1-t^2-1)/(1-t^2)}}} by adding in enough to be able to factorise. Make sure you accept that this line is the same as the previous one.


{{{yb = ((t+1)^2 - t^2 - 1)/(1-t^2)}}}
{{{yb = ((t+1)^2 - t^2 - 1)/((1-t)(1+t))}}}
{{{yb = ((t+1) - t^2 - 1)/(1-t)}}}
{{{yb = (t - t^2)/(1-t)}}}
{{{yb = (t(1-t))/(1-t)}}}


Now for the other one:


{{{x/a = (1+t^2)/(1-t^2)}}}
{{{x/a = (1+t^2+2t-2t)/(1-t^2)}}} -- again make sure that you accept that this line is the same as the previous one. Now we can factorise the top and bottom:


{{{x/a = ((1+t)^2-2t)/((1-t)(1+t))}}}
{{{x/a = ((1+t)-2t)/(1-t)}}}
{{{x/a = (1-t)/(1-t)}}}


Sub this into the first one, to give {{{yb = (tx)/(a)}}}, so {{{t = (aby)/x}}}


As for now? who knows :-)...put this into the 2 originals and you have no t.



jon.