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I am unable to see how to eliminate $t$. Wolfram Alpha fails at it too.

$$x=2t-4t^3$$ $$y=t^2-3t^4$$

I can guess that the curve is a polynomial equation so in principle I can write this as

$$w_1 x^4 +w_2 x^3 +w_3 x^2 +w_4 x +w_5 y^3 +w_6 y^2 +w_7 y +w_8=0$$

for the powers of $t$ to cancel when plugged in I am determine all these $w's$ by substituting the $t's$ for $x(t)$ and $y(t)$ and setting each of the new coefficients of the $t$'s $C_i(w_1,...w_8)$ to zero individually. I have my curve determined if I put each of the $C_i(w_1,...w_8)=0$ and solve for the $w$'s.

All this was in principle, but this is like an examination question so there must be a clever manipulation/way, which I have been unable to find.

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Since I know the source of the problem, and there is no obvious way to eliminate $t$... may I suggest creating a table and plotting points? As a polynomial, we know it will act somewhat predictably, so it shouldn't take too many points to understand what the graph "looks like." It does say "sketch," after all ;) – Tyler Mar 18 '11 at 7:02
Lol! That's a good idea. Ill try to plot a few points and get a good polynomial to fit. Btw, after seeing the actual answer I realize that the method ive posted is also wrong for obvious reasons. – Please Delete Account Mar 18 '11 at 7:56
up vote 8 down vote accepted

My opinion is that you should not try to eliminate $t$ at all. Instead, you should think about how $x$ and $y$ behave as $t$ varies, find some important points such as the critical points with respect to $x$ and $y$, and try to sketch the graph from that information.

$x=2t-4t^3$ means that as $t$ goes from $-\infty$ to $\infty$, $x$ decreases from $\infty$ to a local minimum of $-\sqrt{8/27}$ at $t = -1/\sqrt{6}$, then rises to $\sqrt{8/27}$ at $t = 1/\sqrt{6}$, then falls again to $-\infty$.

$y=t^2-3t^4$ means that $y$ starts and ends at $-\infty$, with local maxima of $1/12$ when $t = \pm 1/\sqrt{6}$, and a local minimum of $0$ at $t = 0$.

Interestingly, $t = \pm 1/\sqrt{6}$ is a critical point for both $x$ and $y$, both of whose derivatives change sign, so the curve forms a cusp at that point.

That's enough information to follow the curve as $t$ goes from $-\infty$ to $\infty$: it comes in from the bottom right quadrant $(\infty,-\infty)$, goes upward and leftward till it hits the cusp $(-\sqrt{8/27},1/12)$, upon which it turns around, passing smoothly through $(0,0)$ where its tangent is horizontal, until it reaches the second cusp $(\sqrt{8/27},1/12)$, and then turns around a second time and exits at the bottom left towards $(-\infty,-\infty)$.

This is consistent with the plot arising from Juan Joder's answer.

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I wouldn't know how to do it in an exam, but using a Gröbner basis routine such as that in Mathematica easily gives the implicit Cartesian equation


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In Mathematica: GroebnerBasis[{x == 2t - 4t^3, y == t^2 - 3t^4}, {x, y}, t] – Juan Joder Mar 18 '11 at 6:57
I googled for the grobner basis algorithm but I got a lot of stuff about polynomial rings etc which I have no background on. Do you know if there is preferrably a method or a specific algorithm for it that I can use to atleast get approximate numerical values? – Please Delete Account Mar 18 '11 at 7:02
Well, internally most computing systems like Mathematica or Maple use Gröbner bases to eliminate variables anyway. I believe there were a few questions on this asked on this very site, but I can't be bothered to search. As for getting numerical values, you're lucky you already have a parametric equation to begin with; the problem of turning an implicit Cartesian equation into a (practical!) set of parametric equations is quite harder. – Juan Joder Mar 18 '11 at 7:06
As for easing the task of sketching, note that $y(t)$ is an even function; it stands to reason that your graph ought to be symmetric about the y-axis (or, as soon as you eliminate the parameter $t$, $x$ will always occur as even powers). Use that to simplify the task of sketching. – Juan Joder Mar 18 '11 at 7:19

Natural points for sketching include: when $x=0$, giving $t = 0 \text{ or } \pm 1/\sqrt{2}$ and so $y = 0 \text{ or } -1/4$; and when $y=0$, giving $t = 0 \text{ or } \pm 1/\sqrt{3}$ and so $x = 0 \text{ or } \pm 2/\sqrt{27}$. So these four points suggest something interesting is going on near the origin.

Just drawing the curve parametrically using Excel gives something like large scale curve

but zooming in gives the more interesting enter image description here

largely matching Rahul Narain's answer and Wolfram Alpha

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You may have a look at Gibson's book "Elementary Geometry of Algebraic Curves", in particular discussions on page 9 and page 178.

As to your problem at hand, you just consider x and y as parameters in equations $f(t) =-4t^3 +2t -x =0$ and $g(t) =-3t^4 +t^2 -y =0$. When x and y are chosen properly there is a value of t solving both equations. That means there is a common root for both polyomials. So to find the relationship between x and y you need only to solve the "resultant" of these two polynomials f(t) and g(t). This resultant is a determinat involving x and y which vanishes if and only if there is such a common root. Hence the resultant $R(f,g)=0$ gives the desired equation relating x and y.

As to your original method, you overlooked the possibility that the equation relating x and y might have monomials like $xy$ and $x^2y$. Thus it is destined to fail.

If my computation is correct, the final equation should be $$\frac{27}{64}x^4-\frac{1}{16}x^2-\frac{9}{2}x^2y+4y(y+\frac{1}{4})^2=0.$$

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