D S
Posted on Thursday, 27 February, 2003 - 11:54 pm:
This is a (pure maths)question from an old longman A level revision guide

The points P(ap2 ,2ap)and Q(aq2 ,2aq) lie on the parabola with equation y2 = 4ax meet at the R [apq,a(p+q)]show the area of triangle PQR is

0.5a (p-q)3

the (P-q) is in a modulus sign
Philip Ellison
Posted on Friday, 28 February, 2003 - 04:52 pm:
There are a number of ways of solving this, but most are very long-winded. Have you found any yet? How much work with vectors have you done?
D S
Posted on Saturday, 01 March, 2003 - 07:52 pm:
philip- i cant think how to solve it- what would be the simplest way?
i thought of using vectors -but want sure how 2
Philip Ellison
Posted on Monday, 03 March, 2003 - 05:32 pm:
If you've met the cross product, then you can use the fact that the area of a triangle P Q , where Q R=a and Q P=b, is 1/2|a×b|. If you haven't learnt about this yet, then I'll be happy to explain.

The next shortest solution I've found is as follows (tho' it may be a bit confusing without the diagram!):

gradient of P R=1/p=tan x

gradient of Q R=1/q = tan y

therefore tan(p-y)=(-tan y+tanp)/(1+tan y tan p)=-1/q

this gives:

area of triangle P Q R=1/2 P R×P Q sin R, where R=((p-y)+x)

we also have:

tan2 x=1/p2

sec2 x=(1+p2)/p2

hence:

cos2 x=p2/(1+p2)

sin2 x=1/(1+p2)

also:

cos2(p-y)=q2/(1+q2)

sin2(p-y)=1/(1+q2)

as x and p-y are both acute:

sin R=(p-q)/(1+p2)1/2(1+q2)1/2

also, P R=a|p-q|(p2+1)

and, Q R=a|p-q|(q2+1)

therefore, the area of triangle P Q R, which is the product of sin R, P and Q is equal to 1/2 a2 |p-q|3, as required.

If you don't follow any of the steps, then please ask, as I've skipped a few stages of working to save time (laziness!).

D S
Posted on Monday, 03 March, 2003 - 10:49 pm:
thanx so much,
i've learnt the cross product and thought of using it- but how exactly would u do it ?
b/c you'd get i's and j's
-sorry 4 being so thick!
Philip Ellison
Posted on Tuesday, 04 March, 2003 - 06:14 pm:
(a1 i +b1 j +c1 k )x(a2 i +b2 j +c2 k )=det(i j k ,a1 b1 c1 ,a2 b2 c2 ), where a comma denotes a change in line (this is a 3 by 3 matrix... could anyone elighten me on how to do these properly?).

You can see this by calculating both the cross product and the determinant of the matrix and rearranging. Once you've got this step, you just need to represent QR and QP as vectors, and work out the determinant of the matrix (with c1 and c2 equal to 0).