Hi
Let A(0,1)
We know that and are colinear
coordinates are (2p,p²-1)
coordinates are (2p-2q,p²-q²)
Therefore 2p(p²-q²)-(2p-2q)(p²-1)=0
Spoiler:
The straight line joining the points P(2p,p²) and Q(2q,q²) passes through the point (0,1)
a) i) Show that pq = -1
ii) Deduce the coordinates of R, the midpoint of PQ, are
(p - 1/p , 1/2(p² + 1/p²))
b) Find the cartesian equation of the locus of R as p varies, giving your answer in the form y=f(x)
OK i have attempted a lame start.
x=2p y=p²
p=x/2 y=(x/2)²
and now i'm stuck
assuming that P and Q are distinct, we must have now look at the slopes: which after simplifying gives you: and so
use the formula you have for the midpoint coordinates and then using the first part of your problem put
ii) Deduce the coordinates of R, the midpoint of PQ, are (p - 1/p , 1/2(p² + 1/p²))
you don't need to do anything else! your answer, is corretc.
b) Find the cartesian equation of the locus of R as p varies, giving your answer in the form y=f(x)
OK i have attempted a lame start.
x=2p y=p²
p=x/2 y=(x/2)²
and now i'm stuck
Hi thanks a lot for your help. I'm still a bit unsure how you derive this, I think i may just be having a mental block!
(2p,p²-1) and (2q,q²-1)
Now I realise this comes from the point A(0,1) but im not sure how
I'm also unsure how this is derived :/
sorry if this is really simple math think i may need to take a break