# Parametric equations: Finding the ordinary equation in $x$ and $y$ by eliminating the parameter from parametric equations

I am having difficult time solving the following equation:

Eliminate the parameter from the parametric equations: $x=\frac{3t}{1+t^3}$ and $y = \frac{3t^2}{1+t^3}$ where $t \ne−1$ and hence find an ordinary equation in x and y for this curve. Find the equation of tangent line to this curve at t = 1.

I am not able to create a (ordinary equation/) parametric equation from the two given parameters $x$ and $y$. Creating the tangent line at $t=1$ is not my concern.

My work
Equation for tangent line:
$P_{tamgent}= \frac{ \frac{d}{dx}x_1(1)}{\frac{d}{dx}y_1(1)}(x-x_1(1))+y_1(1)$

when:
$x_1(t)= \frac{3t}{1+t3}$
$y_1(t)=\frac{3t^2}{1+t3}$

Below is a picture of my solution where the:
Red line = the equation for $x$
Blue line = the equation for $y$
Orange line = the parametric equation of $(x,y)$
Green line = the tangent line at $t=1$

Note: I understand how to solve parametric equations,although not in solving parametric equations where $x$ and $y$ have variables that are difficult to solve for.

Please any help would be much appreciated! Thank you for your time.

## 2 Answers

We have $$x=\frac{3t}{1+t^3}\tag 1$$ $$y=\frac{3t^2}{1+t^3}\tag 2$$ Dividing (2) by (1), we get $$\frac{y}{x}=t\iff t=\frac{y}{x}$$ Now setting the value of $t$ in (1), we get $$x=\frac{3\frac{y}{x}}{1+\frac{y^3}{x^3}}$$ $$x=\frac{3\frac{y}{x}}{\frac{x^3+y^3}{x^3}}$$ $$x^3+y^3-3xy=0$$ Hence, the ordinary equation in the cartesian coordinates $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x^3+y^3-3xy=0}}$$

• @asdfasdfasdf, You need the other way round, right? – lab bhattacharjee Aug 12 '15 at 6:52
• @Harish Chandra Rajpoot what do you mean by "the other way round"?? ( $\infty$ thanks for the hasty answer!!, I have been working on solving this equation correctly for 4 hours) – removed account Aug 12 '15 at 6:55
• @Harish Hey there, I am working through your answer and was wonder what exactly is happening in the last few steps. Before simplifying my last step is equal to $x((x^3+y^3)-3yx)=0$. I've worked through your solution multiple times already, but it appears like an $x$ just disappears, or is this intentional? (Or am I mistaken.. its quite late over here) – removed account Aug 12 '15 at 7:28

$$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{\dfrac{(1+t^3)6t-3t^2(3t^2)}{(1+t^3)^2}}{\dfrac{(1+t^3)3-3t^2(3t)}{(1+t^3)^2}}=\cdots$$

$$\dfrac{dy}{dx}_{(\text{ at }t=1)}=?$$

and $\dfrac yx=\cdots=t$ replace this value of $t$ in one of the given equations to eliminate $t$

• My apologies, this solution appears to be for solving for the tangent line at t=1, my question is directed at creating a parametric equation from $x$ and $y$. Sorry, I will edit my question to be more clear. – removed account Aug 12 '15 at 6:44
• @asdfasdfasdf, Do you mean from the Ordinary Equation? – lab bhattacharjee Aug 12 '15 at 6:48
• Yes, that's what I intended to write. Sorry. – removed account Aug 12 '15 at 6:51