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# Parametric Curve - P3 watch

1. I was trying to plot a curve for the following parametric equations:

x=3tan t
y=2sec t
(cartestian eqtn is 9y^2 - 4x^2 = 36)

I got stuck on how to calculate what it looked like myself due to difficulties arriving in having imaginary no's (this is P3 so we should need them!!!). So i used my graph drawing program and got the following graph. Should I need to be able to work out such a complicated shape at P3 and can someone explain how to obtain the shape? Thanks
2. (Original post by Hoofbeat)
(cartestian eqtn is 9y^2 - 4x^2 = 36)
does the graph look like this...(for positive y), but i'm not sure what happens below the y-axis, except that maybe cuts at y=-2???
Attached Images

3. For -pi<t<pi it will look like this. It's a hyberbola that has been reflected in the line y=x because the general equations of a hyperbola are y=btant, x=asect, cartesian eqn is x^2/a^2 - y^2/b^2 = 1. I'm not sure how they'd expect you to do this for p3 though - I guess just draw y vs. t and x vs. t and combine them

Edit - First graph had coefficients the wrong way round
Attached Images

4. (Original post by Bezza)
For -2pi<t<2pi it will look like this. It's a hyberbola that has been reflected in the line y=x because the general equations of a hyperbola are y=btant, x=asect, cartesian eqn is x^2/a^2 - y^2/b^2 = 1. I'm not sure how they'd expect you to do this for p3 though - I guess just draw y vs. t and x vs. t and combine them
i wouldn't have thought that you'd get that curve from the cartesian though??
5. (Original post by mockel)
i wouldn't have thought that you'd get that curve from the cartesian though??
If you replace x with (-x) or y with (-y) the equation stays the same so it is symmetrical in both the x and y axes
6. (Original post by Bezza)
If you replace x with (-x) or y with (-y) the equation stays the same so it is symmetrical in both the x and y axes
but how do you know about the asymptotes?

btw, are they called oblique asymptotes when they are like that, and would that be in P3?
7. (Original post by mockel)
but how do you know about the asymptotes?

btw, are they called oblique asymptotes when they are like that, and would that be in P3?
I know about the asymptotes cos you study the hyperbola in p5! If you rearrange the cartesian equation to the standard form and you have y^2/4 - x^2/9 = 1. For large x and y this is close to y^2/4 - x^2/9 = 0, (y/2 + x/3)(y/2 - x/3) = 0 so the asymptotes are y = ± 2x/3. This isn't even needed for p5, you only need to state the asymptotes so don't worry about it too much.

Yes they are and no it wouldn't. It does seem quite a strange p3 question - where did it come from hoofbeat?
8. (Original post by Hoofbeat)
I was trying to plot a curve for the following parametric equations:

x=3tan t
y=2sec t
(cartestian eqtn is 9y^2 - 4x^2 = 36)

I got stuck on how to calculate what it looked like myself due to difficulties arriving in having imaginary no's (this is P3 so we should need them!!!). So i used my graph drawing program and got the following graph. Should I need to be able to work out such a complicated shape at P3 and can someone explain how to obtain the shape? Thanks
surely you just make a table, do some values and plot it?
9. (Original post by Bezza)
I know about the asymptotes cos you study the hyperbola in p5! If you rearrange the cartesian equation to the standard form and you have y^2/4 - x^2/9 = 1. For large x and y this is close to y^2/4 - x^2/9 = 0, (y/2 + x/3)(y/2 - x/3) = 0 so the asymptotes are y = ± 2x/3. This isn't even needed for p5, you only need to state the asymptotes so don't worry about it too much.

Yes they are and no it wouldn't. It does seem quite a strange p3 question - where did it come from hoofbeat?
ok, thanks
10. (Original post by Bezza)
It does seem quite a strange p3 question - where did it come from hoofbeat?
She got it from www.ajimal.com - on the graph sketching exercise. I got stuck on exactly the same question
11. I just looked at the question and it only wants the sketch -pi/2 < t < pi/2 so you only need to draw the top curve since for this range of t, sect is positive
12. (Original post by Bezza)
I just looked at the question and it only wants the sketch -pi/2 < t < pi/2 so you only need to draw the top curve since for this range of t, sect is positive
but you would still have to include those asymptotes....so we wouldn't get anything like this for p3 (i.e. no oblique asymptotes)
13. (Original post by mockel)
but you would still have to include those asymptotes....so we wouldn't get anything like this for p3 (i.e. no oblique asymptotes)
Yeah, the asymptotes would still be needed for a proper sketch of the curve. If something like this did come up in a p3 exam you certainly wouldn't need to put the asymptotes for full marks. It's only a sketch so they probably only want - crosses y-axis at (0, 2), symmetrical in y-axis, general shape correct
14. Thanks Bezza - that's the shape graph I got when I used my graph drawing program but it didnt let me attach it on here! grrrr

Actually Sally it came from the Heinman Edexcel P3 Revision Guide (Pg 21 Q9) but I couldn't see a nice way to work out the shape of the graph without plotting points! Glad to hear you don't think it'll come up in the P3 paper, if it does I'll just try values and plot the graph rather than just "sketch". Looks like this will be the sort of stuff I'll be teaching myself over the summer hols!
15. (Original post by Hoofbeat)
Thanks Bezza - that's the shape graph I got when I used my graph drawing program but it didnt let me attach it on here! grrrr
That's ok. I managed to find a free graph drawing program called graphcalc which does normal, parametric and polar graphs, and also has a button to copy the graph straight to the clipboard which is useful.
16. [QUOTE=Hoofbeat]
Actually Sally it came from the Heinman Edexcel P3 Revision Guide (Pg 21 Q9) but I couldn't see a nice way to work out the shape of the graph without plotting points! QUOTE]

Well its simliar to the question i showed Mr. V - he just said that if we get sthg as hard as that, just plot in some values
Chloé - MSN NOW!!!

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