Question

Find parametric equations for the rectangular equation.

1) y = 3x - 4

2) (x-2)2 + (y-4)2 = 4
(I believe the outside 2s are meant to be ^2). I know how to find a rectangular equation for parametric equations, but not the other way around, please help.

Answer 1

first one, though it looks like cheating, you could say \(x=t, y = 3t-4\)

Answer 2

second one is a circle with center \((2,4)\) and radius \(2\)

Answer 3

i think it is just
\[x=2+2\cos(t), y = 4+2\sin(t)\] if i remember correctly

Answer 4

re: the first one, that's the answer I got, however would that answer be acceptable if it's mandatory to show work?

Answer 5

sure what else can you do? it is a line, y is written explicitly in terms of x

Answer 6

you need work for the second one too?

Answer 7

okay good point, just am worried because my school is very particular about showing work so I even am going to have to explain that's its' because its a line in order for points to be given.

and yes. :)

Answer 8

hmmm i was hoping you would say no
we can start with
\[x=\cos(t), y = \sin(t)\] which is the well known unit circle
to make the radius 2 it is therefore
\[x=2\cos(t), y = 2\sin(t)\] now translate so that the center is \((2,4)\)

Answer 9

that gives
\[x=2+2\cos(t), y = 4+2\sin(t)\]
if there is some algebra way i am afraid i do not know it

Answer 10

in fact, i would say there is not
it is the translation of the unit circle

Answer 11

You can certainly prove it algebraically, though:

Replace x with 2+2cos(t), and y with 5 + 2sin(t), in the equation (x-2)^2 + (y-4)^2 = 4

You will get 4cos^2(t) + 4sin^2(t) = 4(cos^2(t) + sin^2(t)) = 4(1) = 4

Answer 12

@satellite73

Answer 13

yes sure but that is working backwards

Answer 14

Agreed. It's just a nice check, in case one is not sure.

Answer 15

thank you so much, both of you are lifesavers. :)