Elimination of $'t'$ in a locus problem

Question

Okay , so I need to find a relation between $x$ and $y$ using

$$x=\dfrac{3+3t^2}{t^2-1} \qquad and \qquad y=\dfrac{3t}{t^2-1}$$

I tried by solving for $t$ from $y$ , but that ended up in a quadratic in $t$ .Now solving for $t$ and then putting $t$ in expression of $x$ further complicates the result.

Isn't there another , less tedious way ? Any help would be appreciated !

Answer 1

Hint:

Use $$(1-t^2)^2+(2t)^2=(1+t^2)^2\iff\left(\dfrac{1-t^2}{1+t^2}\right)^2+\left(\dfrac{2t}{1+t^2}\right)^2=1$$

or put $t=\tan u$

Answer 2

Hint... Consider that $$\frac xy=\frac 1t+t$$ and $$\frac 3y=t-\frac 1t$$ so add these equations and you can get $t$ which can then be substituted