1. ## Parametrics Help

Hello,

If ya havent figured out by my post number, I am new to the forum. I have been out of the math game for a while and limping through Calc I and II. Since, I am taking Calc II currently via distance learning, it makes it tough sometimes to understand the material. So here is my first question I need help with:

I need to sketch the curve represented by the equation and write the corresponding rectangular equation (which I get generally):

x=t^2+t and y=t^2-t

I automatically tried solving x in terms of t but that led me to t=sqrt(x-t)

I dunno

2. Originally Posted by ajwilkes
Hello,

If ya havent figured out by my post number, I am new to the forum. I have been out of the math game for a while and limping through Calc I and II. Since, I am taking Calc II currently via distance learning, it makes it tough sometimes to understand the material. So here is my first question I need help with:

I need to sketch the curve represented by the equation and write the corresponding rectangular equation (which I get generally):

x=t^2+t and y=t^2-t

I automatically tried solving x in terms of t but that led me to t=sqrt(x-t)

I dunno
If you want to solve for t interms of x, write the equation as t^2+ t- x= 0 and use the quadratic formula: $t= \frac{-1\pm \sqrt{1+ 4x}}{2}$.

But I think the simplest thing to do is to subtract both equations to eliminate $t^2$ and get x- y= 2t so t= (x- y)/2. Now [tex]t^2= (x-y)^2/4[tex] so $x= t^2+ t= (x-y)^2/4+ (x-y)/2= (x-y)((x-y)/4+ 1/2)$ $= (1/4)(x-y)(x- y+ 2)$

$x^2- 2xy+ y^2+ 2x- 2y= 4x$

Looks like a tilted parabola to me.

3. Thank you for the help.