| Posted on Sunday, 04 January, 2004 - 09:25 pm: |
Can you explain how i should go about this question please?
Sketch the conic with polar equation a/r=1+cos(theta)
| Posted on Sunday, 04 January, 2004 - 10:20 pm: |
Here is a quick start: find the values of a for which the curve is symmetrical. This is when f(2a-q)=f(q). Also find if there are any tangents parallel to the x and y-axis respectively. Note that there are no tangents to the pole. Mark
| Posted on Sunday, 04 January, 2004 - 10:28 pm: |
Oliver, do you recognise the equation r=l/(1+e×cosq)? I should have thought that the equation is quoted in the text/formula book.
| Posted on Sunday, 04 January, 2004 - 10:29 pm: |
You can cheat here and change the equation into Cartesian (multiply by r, then play around a bit using the facts that x = rcosq and r2 = x2 + y2 ) or you can just take a moment to think about how r changes as theta increases (the fact that as theta goes to pi, r shoots to infinity should immediately tell you which one of the three conics this is).
Also (although don't quote me on this one) I think an A-Level result is that a/r = 1 + e cos theta represents a conic of eccentricity e (in this case e = 1 telling you it's a *ar*b*l*)
Marcos
| Posted on Sunday, 04 January, 2004 - 10:40 pm: |
sorry Mark, i don't understand the usage of 'a'.
Angelina, i haven't been given that formula in my text book
I need to know how to this without going through cartesian so that i can apply it to all questions like this
| Posted on Sunday, 04 January, 2004 - 10:52 pm: |
Okay, here's what I'd do (i.e. not necessarily the best way!): Note that it's symmetrical about the line q = 0 and that as q goes to p, the RHS goes to 0 so that r shoots to infinity. This basically tells us that it's a parabola (since we're given that it's a conic) symmetrical about the initial line spreading out towards the 2nd/3rd quadrants. This is enough to sketch it, also helpful is where it cuts the axis, found by putting in q = 0 (comes out to be a/2 along). And we're done! (If you want you may want to add a couple more points, say where q = ±p/2.) Marcos
| Posted on Sunday, 04 January, 2004 - 11:05 pm: |
Well, going through cartesian gives you the general equation which is what I quoted above.
Or you can work it out from the first principles, which usually helps you remember it a lot better:
Remember that a conic is a locus of point in a place such that its distance from a fixed point S is a constant multiple of its distance. Let r be the distance from the fixed point, e the constant multiplier (eccentricity), l the length of semi-latus rectum (as shown in the figure).
Now P lies on the conic, herefore:
SP = e*PM (according to definitions of conics)
< => r = e*(SX - r*cos q)
< => r*(1 + e*cos q) = e*SX
Now when q = pi/2, e*SX = r, which is PN, i.e. l .
This gives the equation l/r = 1 + e*cos q, and you should now be able to sketch the graph.
| Posted on Sunday, 04 January, 2004 - 11:10 pm: |
Angelina, is 'l' replaced by 'a' in this case then?
| Posted on Sunday, 04 January, 2004 - 11:12 pm: |
Yes