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.
Best Answer
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 $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x^3+y^3-3xy=0}}$$