"Find the implicit form of the curve defined by parametric equations $x = t+1,y=\frac{1}{t^{2}}$"
How can I clear $t$ to arrive at the implicit equation?
[Math] Parametric to implicit form of a curve
parametric
Related Solutions
Another way:
we have $x=2\cos t-\cos2t\ \ \ \ (1)$
$\implies x=2\cos t-(2\cos^2t-1)\iff x-1=2\cos t(1-\cos t)$
and $y=2\sin t-\sin2t\ \ \ \ (2)$
$\implies y=2\sin t -2\sin t\cos t=2\sin t(1-\cos t)$
On division, $\displaystyle\frac yx=\tan t$
$\displaystyle\implies \cos2t=\frac{1-\tan^2t}{1+\tan^2t}=\frac{x^2-y^2}{x^2+y^2}$
and $\displaystyle\cos t=\frac1{\sec t}=\pm\frac1{\sqrt{1+\tan^2t}}=\frac{\pm x}{\sqrt{x^2+y^2}}$
Put these values in $(1)$ and square to remove $\pm$
In Macaulay2
R=QQ[s,t,x,y,z,MonomialOrder=>Eliminate 2]
I=ideal(x-2*s*t*(3*t^4+50*t^2*s^2-33*s^4),y-2*(7*t^6-60*t^4*s^2+15*t^2*s^4+2*s^6),z-(t^2+s^2)^3)
gens gb I
yields a term free of $s, t$:
$625x^6+1875x^4y^2+1875x^2y^4+625y^6-182250x^4yz+364500x^2y^3z-36450y^5z+585816x^4z^2+1171632x^2y^2z^2+585816y^4z^2-41620992x^2z^4-41620992y^2z^4+550731776z^6$
Dehomogenising $z$ we get:
$625x^6+1875x^4y^2+1875x^2y^4+625y^6-182250x^4y+364500x^2y^3-36450y^5+585816x^4+1171632x^2y^2+585816y^4-41620992x^2-41620992y^2+550731776=0$
Which plots to the same as your parameterization
Edit
I didn't have to use the homogenised versions of everything, because the more straightforward clearing of denominators (see exercises 3.3.13-14 of Ideals, Varieties and Algorithms)
R=QQ[t,x,y]
I=ideal((t^2+1)^3*x-2*t*(3*t^4+50*t^2-33),(t^2+1)^3*y-2*(7*t^6-60*t^4+15*t^2+2))
gens gb I
gives the relation directly.
Interesting (to me at least) is that the original $x=−9\sin(2t)−5\sin(3t), y=9\cos(2t)−5\cos(3t)$ also satisfies this relation.
Edit2
Maple does it too:
> with(Groebner);
> Basis({(t^2+1)^3*x-2*t*(3*t^4+50*t^2-33), (t^2+1)^3*y-2*(7*t^6-60*t^4+15*t^2+2)}, tdeg(t, x, y));
Best Answer
HINT:
If $x=\dfrac1{t+1}\iff t+1=\dfrac1x\iff t=\cdots$
If $x=t+1\iff t=x-1$
Put this value in $y=\dfrac1{t^2}$
Alternatively, $y=\dfrac1{t^2}\iff t^2=\dfrac1y\ \ \ \ (1)$
and $x=t+1\iff t=x-1\implies t^2=(x-1)^2\ \ \ \ (2)$
Compare the values of $t^2$ in $(1),(2)$ to eliminate $t$