Answer to Question #175731 in Calculus for Shahir Sheikh

Question

Using a table, sketch the parametric equations, including the orientation arrows.

x = t^2 -2t

y = -t² + 4t

0 ≤ t ≤ 3

Answer 1

$x=t^2-2t\\y=-t^2+4t$

$0\leq t\leq3$

Graph of these two parametric equations are

Now upon solving the parametric equations we get,

$t^2=x+2t\\and\ \ \ t^2=4t-y$

Equating the above two equations:

$x+2t=4t-y\\t=\dfrac{x+y}{2}$

Putting the value of t in $x=t^2-2t$

$x=\dfrac{(x+y)^2}{4}-2\dfrac{(x+y)}{2}$

on solving,

$x^2+y^2-8x-4y+2xy=0$

Table:

and final locus of the parametric equation will be: