The Student Room Group

Parametric Equations(Trig)

Find the Cartesian equation of the curve given by the following parametric equation:

x= 2 sin t-1, y=5cos t+4, 0<t<2pi

I understand that this can be done using substitution and utilising well-known trig identities. I ended up getting an equation for y, as follows:

y=195x2 y= \frac{19-5x}{2}

However, when I checked the solution, I was disappointed to find that the intended solution would be in the form of an equation for a circle. What I want to know is: what, in the question, implied that this is what I should have done, rather than write a cartesian equation in the form "y=..."? I would hate to make this mistake in an exam, so if anyone could point out my mistake(s), I would really appreciate it :smile:
Original post by Illidan2
Find the Cartesian equation of the curve given by the following parametric equation:

x= 2 sin t-1, y=5cos t+4, 0<t<2pi

I understand that this can be done using substitution and utilising well-known trig identities. I ended up getting an equation for y, as follows:

y=195x2 y= \frac{19-5x}{2}

However, when I checked the solution, I was disappointed to find that the intended solution would be in the form of an equation for a circle. What I want to know is: what, in the question, implied that this is what I should have done, rather than write a cartesian equation in the form "y=..."? I would hate to make this mistake in an exam, so if anyone could point out my mistake(s), I would really appreciate it :smile:


Post your working in order for us to point out your mistakes.

Secondly, you ended up with an equation of a line, but I'm not sure how this isn't obviously wrong because x,y are non-linear functions which change at different rates.

To answer your question, I don't know what you mean. It's asking for the Cartesian equation so the circle's equation is perfectly fine.

You should instead had noticed that you can isolate sint\sin t and cost\cos t in their respectively equations, square those equations, and add them to make 1 so that there is no tt.
Reply 2
I tried to post some images of my working in my initial post, but there appeared to be an error in the upload so I was unable to do so. Here is an imgur link to my working(I apologise for the presentation). https://imgur.com/a/tvxPU

I knew I needed to eliminate t, but I didn't realise what I should have been aiming for was the equation of a circle. I found an equation for sin^2t, and then used the identity sin^2t+cos^2t=1, in a sense, to eliminate t from y and replace it with x.
(edited 5 years ago)
Original post by Illidan2
I tried to post some images of my working in my initial post, but there appeared to be an error in the upload so I was unable to do so. Here is an imgur link to my working(I apologise for the presentation). https://imgur.com/a/tvxPU

I knew I needed to eliminate t, but I didn't realise what I should have been aiming for was the equation of a circle. I found an equation for sin^2t, and then used the identity sin^2t+cos^t=1, in a sense, to eliminate t from y and replace it with x.


You are simply making rookie mistakes like saying that (a+b)2=a2+b2(a+b)^2 = a^2+b^2 and a2+b2=a+b\sqrt{a^2+b^2} = a+b which are not true.

Reply 4
I'm confused. I can see what you're saying about rookie mistakes, I understand that (a+b)2 (a+b)^2 is not equal to a2+b2 a^2+b^2 and so forth. I do sometimes forget and make a mistake like that, but I don't understand how i've done it here. I can see an error on the second line you've marked as incorrect, but I don't spot an error with the first incorrect line. Could you explain?
Original post by Illidan2
I'm confused. I can see what you're saying about rookie mistakes, I understand that (a+b)2 (a+b)^2 is not equal to a2+b2 a^2+b^2 and so forth. I do sometimes forget and make a mistake like that, but I don't understand how i've done it here. I can see an error on the second line you've marked as incorrect, but I don't spot an error with the first incorrect line. Could you explain?


That comes from squaring y=5cost+4y=5\cos t + 4 which leads to y2=(5cost+4)2y^2 = (5 \cos t + 4)^2 which you said is 25cos2t+1625\cos^2 t + 16 so this is just applying the incorrect result of (a+b)2=a2+b2(a+b)^2 = a^2+b^2.
Reply 6
Ahh. That would yield the quadratic in the form 25cos2t+40cost+16 25cos^2t +40cost + 16 . That would explain how I got a linear solution when I shouldn't have done.
(edited 5 years ago)
Original post by Illidan2
Ahh. That would yield the quadratic in the form 5cos2t+40cost+16 5cos^2t +40cost + 16 . That would explain how I got a linear solution when I shouldn't have done.


You found that sin2t=(x+12)2\sin^2 t = \left( \dfrac{x+1}{2} \right)^2 which is fine, leave it.
Now do the same procedure to get cos2t=...\cos^2 t = ...

Add the two equations and use sin2t+cos2t1\sin^2 t + \cos^2 t \equiv 1 which would then be the correct approach and you end up with a Cartesian eq.

P.S. This is not an equation of a circle like you said, it's an equation for an ellipse.
(edited 5 years ago)
Reply 8
Damn. That was a lot simpler than I thought it was. I originally took a pretty long-winded way to find a Cartesian equation, when all I had to do was find two equations and use one trig identity. Thank you so much again, as always! :smile:
(edited 5 years ago)

Quick Reply

Latest