Maths: parametric eqns of a parabola

[Peter Pedant]

A variable chord of a parabola y2=4ax is of constant length 2a Prove that its midpoint lies on a curve given by (y2-4ax)(y2+4a2)+4a2=0. [S level 1965)

[Peter Pedant]

yup it is proper maths time, and just in case you say ask your teacher, I dont have one.

I have got this far: two points: (ap2 2ap) and (aq2 2aq)
The mid pt is gonna be: [a(p2-q2)/2, a(p-q) ]

and for locus q I need to solve x = a(p2-q2)/2 , y =a(p-q)

and the constant length condition is edpressed as: a2(q2-p2)2 + 4a2(p-q)2= 4a2

and I just cant get it out. (the solution that is)


I am crezzy man: I need help.

well its better than how do you do simulaneous equations innit ?
any pointers gratefully recd.

[vascop]

First thing to say is that the mid point is NOT what you say in your post, it should be:
[a(p2+q2)/2,a(p+q)]
Think of x=6 and x=10, the mid point is (6+10)/2=8.
Your formula gives half the distance between the two points in x and y directions

[vascop]

I have now solved this.
Since the chord through x=a/4 at right angles to the x axis is of length 2a then the point (a/4,0) must be a point on the locus of the mid points. So substituting this point into your curve seems to indicate that the last term should be 4a4 NOT 4a2. So it should be:

(y2-4ax)(y2+4a2)+4a4=0 (1)

Now taking as before x=a(p2+q2)/2 and y=a(p+q) and substituting into (1) above leads to:
(p2-q2)2+4(p-q)2-4=0

Now this is exactly the condition for the chord to be of length 2a. So we have proved that
(y2-4ax)(y2+4a2)+4a4=0 is the locus of the mid points of a chord of length 2a.

[Peter Pedant]

.

Oh God ! Vascop - you little Hero !

thanks - just got in fr London
and will settle dn and read the answer tomorrow

very many thanks