Parabola (AKA The Devil) And Integration (AKA Devil Spawn) (1 Viewer)

[mazza_728]

Hey guys
im having huge trouble in both topics the parabola and integration.
Ive always understood things like im not braggin, well i dont mean to but ive always been able to understand all topics in all of my subjects.
I cannot get these! with the parabola im having huge problems interpreting the question, realising wat the graph is and knowing wat i have to do.
e.g. P is a given point with paramater p on the parabola x²=4ay, with focus S. A ine is drawn from S, perpendicular to SP and meetsthe normal at P in the poin Q. PN, QM are perpendicular to the axis of the parabola.
a) Find the ordinate at Q
b) If D is the intersection of the directrix and the axis of the parabola, prove that DM=2DN.

e.g.2 PQ is a focal chord in the parabola x²=4ay
a) PT is drawn parallel to the tangent at Q and QT is drawn parallel to the tangent at P. Show that the locus of T is x²=a(y-3a)
b) If M is the mid-point of the focal chord PQ and a line through M, parallel to the axis of the parabola, meets the normal at P in A, find the locus of A.

Even when typing out those questions i forgot how to find the locus!!! what is wrong wit me? is anyone finding the same prob??

With integration well truthfully i wouldnt have a clue wat im doing. this could be due to the fact that i havent had any real previous practice, we havent done it in 2 unit and he is really rushing thru the topic. im not sure wat i have to do and am getting really confused - do i integrate? differentiate? find the primitave?? i dunno - we using subsitution at the moment and its just not working for me!
e.g. (1-sin²3t).cos3t where u = sin3t

Thanks guys any advice tips criticism anything would help.
i feel sooo stupid!Why cant i do this???

 

[Heinz]

I hate parametrics so ill leave that for someone else. In finding the locus, just equate the x and y ordinates to find some sort of relationship. As for integration (something i can do)
(1-sin^2(3t).cos3t where u = sin3t. du/dt = 3cos3t
Therefore, the integral of (1-sin23t).cos3t = 1/3(1-u^2)du =u/3 -(u^3)/9 + c = sin3t/3-sin^3(3t)/9 + c

 

[~*HSC 4 life*~]

Originally posted by mazza_728

what is wrong wit me? is anyone finding the same prob??


i feel sooo stupid!Why cant i do this???

don't worry gal, parametrics is a bitch
-well for me it is as well, i think, as well as atempting questions, you need to see a lot of the typical exams questions and solutions, the parametric questions i have seen so far in exam papers ahev all been pretty similar.quite stantard
its just so trciky because you have to udnerstand and construct diagrams with the instructions given and also derive everything...

just know that THERE ARE MANY OUT THERE THAT FEEL THE SAME WAY WE DO!

apparently, well so i have heard, parametrics is one of the hardest 3u topics (??) so yeah, don't worry too much because chances are if you find it hard then others will too

 

[CM_Tutor]

Parametrics is not too bad so long as you remember the golden rules:

Rule 1: Draw a diagram. (Make it big, that way you can draw onto it, and your marker can see what you've done. Any diagram that can be hidden under a postage stamp is TOO SMALL!)

Rule 2: Make your diagram approximately to scale. This is because if you misinterpret the question, a diagram roughly to scale will usually make it obvious (by looking ridiculous).

Rule 3: This is mostly fancy co-ordinate geometry, so expect to be using co-ordinate geometry methods (distance formula, point-gradient form of a line, etc.) with some extra stuff thrown in (like calculus and plane geometry).

Rule 4: Don't let the algebra phase you. This stuff isn't difficult provided you have a clear idea of what you are trying to achieve, but it is often fiddly. (Extension 2 people, this piece of advice goes double for conics!)

Let's illustrate with one of your examples:

P is a given point with paramater p on the parabola x2=4ay, with focus S. A ine is drawn from S, perpendicular to SP and meetsthe normal at P in the poin Q. PN, QM are perpendicular to the axis of the parabola.
a) Find the ordinate at Q
b) If D is the intersection of the directrix and the axis of the parabola, prove that DM=2DN.

Start with a blank page. Freehand, draw a LARGE parabola, and then draw on x and y axes, with the origin at the vertex of the parabola. (It's much easier to draw a realistic parabola without the axes - try it).

Now, locate the focus S. This at (0, a), and thus lies on the y-axis inside the parabola. DON'T just put it anywhere, as we want to be roughly to scale. Remember that the latus rectum (the focal chord that is perpendicular to the axis) has length 4a. Thus, the focus is a distance up the y-axis such that the width of the parabola at that height is four times the distance above the vertex.

P(2ap, ap^2) - that's what the "parameter p" bit means - is any point on the parabola. So, mark a point P on the parabola, in the first quadrant - I generally go about twice as high as the focus. Join PS.

Now, draw the line through S that is perpendicular to PS.

You are interested in the normal at P, but normals are much harder to draw correctly than are tangents. So, draw in the tangent at P, for a couple of cm on either side of P. The noraml at P can now be drawn by drawing the perpendicular to the tangent you have drawn. This normal should cross the line which is perpendicular to S. Mark this point as Q.

The question then says: "PN, QM are perpendicular to the axis of the parabola". This means that N and M lie on the axis. So, draw a horizontal line from P to the y-axis, meeting the y-axis at N. Similarly, draw a horizontal line from Q to the y-axis, meeting the y-axis at Q.

We now need the directrix. This is the line y = -a, and thus is a horizontal line crossing the y-axis as far below the vertex as the focus was above it. The directrix meets the y-axis at D, according to the question.

If your diagram is drawn correctly, you should have marked on the y-axis (reading from top to bottom) the points M, N, S, O (the origin) and D.

The ultimate goal of the question is to show that DM = 2DN, and this should appear roughly true on your diagram. It should be clear that DM = 2DN actually means that N is the midpoint of DM - This realisation makes the question quite easy, as D is at (0, -a), and N is at (0, ap^2). Thus, all we need to do is prove that M is at (0, a + 2ap^2), or that the y co-ordinate of Q is a + 2ap^2.

Now that we know what to do, let's do it, as follows:

Find the equations of QS (co-ordinate geometry) and PQ (calculus and co-ordinate geometry), and solve them simultaneously for Q.

Working:

m(PS) = (ap^2 - a) / (2ap - 0) = (p^2 - 1) / 2p using the gradient formula.
Since QS is perpendicular to PS, m(QS) = -1 / m(PS) = 2p / (1 - p^2)
Thus QS is y - a = 2p(x - 0) / (1 - p^2)
ie y = a + 2px / (1 - p^2) ......... (1)

Since x^2 = 4ay, y' = x / 2a, and so at P, m(tang) = 2ap / 2a = p
So, at P, m(norm) = -1 / p, and so normal is y - ap^2 = - (x - 2ap) / p
ie x = 2ap + ap^3 - py ......... (2)

Solving (1) and (2) simultaneously will give the co-ordinates of Q, but we're really only interested in the y co-ordinate, so we want to eliminate x.

Putting (2) into (1) and multiplying by (1 - p^2), we get y(1 - p^2) = a(1 - p^2) + 2p(2ap + ap^3 - py)
Expanding and collecting like terms, this becomes y(1 - p^2 + 2p^2) = a(1 - p^2 + 4p^2 + 2p^4)
ie y(1 + p^2) = a(1 + 3p^2 + 2p^4) = a(1 + p^2)(1 + 2p^2)
So, on dividing by (1 + p^2), the y co-ordinate of Q is y = a(1 + 2p^2).

Now, since QM is horizontal, and M lies on the y-axis, M must be (0, a + 2ap^2).
Since PN is horizontal, and N lies on the y-axis, N must be (0, ap^2).
Since D lies on the directrix (y = -a) and the y-axis, D must be (0, -a).

The midpoint of DM is (0, [-a + a + 2ap^2] / 2), ie. (0, ap^2), ie N.

So, since N is the midpoint of DM, DM = 2DN (as required).

 

[Calculon]

Parabola aka the Devil Integration aka Devil Spawn

Newton invented integration
:. Newton is the devil
But there is only one devil
:. Newton is a parabola
How does that work?

 

[siavash_s_s]

u ought to try 4unit conics (evil muawhahahahhaha)

 

[Calculon]

Originally posted by siavash_s_s
u ought to try 4unit conics (evil muawhahahahhaha)

I found parametrics more evil when I first did it cos once you get the concepts down its easy

 

[Grey Council]

wtf, say thank you to CM_Tutor? lol

Thanks a lot for your help, CM. ^__^

 

[Calculon]

LOL, thats the maddest avatar