x Turn on thread page Beta
 You are Here: Home >< Maths

# Parametric equations watch

1. Hopefully this is the right place for this

Hey I would like some help with these questions.

1.) x=t/(2t-1) y=t/(t+3)
2.) x=sin2theta y=costheta.3
.)x=sintheta +2costheta y=2sintheta +costheta

For all I need to find the Cartesian form. For the first one I got it down to y=x/(7x-3) and for the second one I got it to be x^2=y^2(4-4y^2). As for the third one I have no answer
2. (Original post by Sylarus)
Hopefully this is the right place for this

Hey I would like some help with these questions.

1.) x=t/(2t-1) y=t/(t+3)
2.) x=sin2theta y=costheta.3
.)x=sintheta +2costheta y=2sintheta +costheta

For all I need to find the Cartesian form. For the first one I got it down to y=x/(7x-3) and for the second one I got it to be x^2=y^2(4-4y^2). As for the third one I have no answer
1) and 2) are correct.

For 3), try squaring x and y and also consider their product.
3. (Original post by Sylarus)
Hopefully this is the right place for this

Hey I would like some help with these questions.

1.) x=t/(2t-1) y=t/(t+3)
2.) x=sin2theta y=costheta.3
.)x=sintheta +2costheta y=2sintheta +costheta

For all I need to find the Cartesian form. For the first one I got it down to y=x/(7x-3) and for the second one I got it to be x^2=y^2(4-4y^2). As for the third one I have no answer

For the third one, obtain expressions for and , in terms of x,y, and use a well known identity to eliminate theta.

Edit: Nothing for 50 minutes and then two at once!
4. If you are saying use cos squared theta add sign squared theta = 1 I've tried using it like this.

Costheta= y-2sintheta
Sintheta= x-2costheta

Then squaring I got this big messy expansion which I can't seem to get the Thetas out of
5. (Original post by Sylarus)
If you are saying use cos squared theta add sign squared theta = 1 I've tried using it like this.

Costheta= y-2sintheta
Sintheta= x-2costheta

Then squaring I got this big messy expansion which I can't seem to get the Thetas out of
Ghostwalker may have to advise you here because I'm not sure.

An alternative method (as mentioned in my last post) : Compare and .
6. (Original post by Sylarus)
If you are saying use cos squared theta add sign squared theta = 1 I've tried using it like this.

Costheta= y-2sintheta
Sintheta= x-2costheta

Then squaring I got this big messy expansion which I can't seem to get the Thetas out of
You haven't got cos/sin in terms of x,y.

Using the two equations, you can eliminate sin, and get cos as a function of x,y.
Similarly you can eliminate cos, and get sin as a function of x,y.

Then apply the identity.
7. (Original post by ghostwalker)
You haven't got cos/sin in terms of x,y.

Using the two equations, you can eliminate sin, and get cos as a function of x,y.
Similarly you can eliminate cos, and get sin as a function of x,y.

Then apply the identity.
Yes that's a much better way. I should have spotted it.
8. K so after some rearranging got it to (2x-y)^2+(2y-x)^2=1
9. (Original post by Sylarus)
K so after some rearranging got it to (2x-y)^2+(2y-x)^2=1
Close. But it doesn't equal 1.

As a check you can choose a value for the parameter, theta = 0 say. Sub into your original equations, and check if the x,y values that come out satisfy your cartesian equation.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 4, 2015
Today on TSR

### University rankings 2019

Cambridge at number one

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams