1. ## Loci...?

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

2. Hi

Let A(0,1)
We know that $\overrightarrow{AP}$ and $\overrightarrow{QP}$ are colinear

$\overrightarrow{AP}$ coordinates are (2p,p²-1)
$\overrightarrow{QP}$ coordinates are (2p-2q,p²-q²)

Therefore 2p(p²-q²)-(2p-2q)(p²-1)=0

Spoiler:

-2pq²+2qp²+2p-2q=0
pq(-q+p)+(p-q)=0
(p-q)(pq+1)=0
Since p and q are different => pq=-1

3. Originally Posted by djmccabie

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
assuming that P and Q are distinct, we must have $p \neq q \neq 0.$ now look at the slopes: $\frac{p^2 - 1}{2p} = \frac{q^2 - 1}{2q},$ which after simplifying gives you: $(pq + 1)(q-p)=0$ and so $pq=-1.$

ii) Deduce the coordinates of R, the midpoint of PQ, are (p - 1/p , 1/2(p² + 1/p²))
use the formula you have for the midpoint coordinates and then using the first part of your problem put $q=\frac{-1}{p}.$

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
you don't need to do anything else! your answer, $y=\frac{x^2}{4},$ is corretc.

4. Originally Posted by NonCommAlg
assuming that P and Q are distinct, we must have $p \neq q \neq 0.$ now look at the slopes: $\frac{p^2 - 1}{2p} = \frac{q^2 - 1}{2q},$ which after simplifying gives you: $(pq + 1)(q-p)=0$ and so $pq=-1.$
Thanks very much for your help I'm still a bit confused. How did you deduce (p^2-1)/2p = (q^2-1)/2q

Also i can't see how it simplifies to that could you add each step please its killing me

5. Originally Posted by djmccabie
Thanks very much for your help I'm still a bit confused. How did you deduce (p^2-1)/2p = (q^2-1)/2q

Also i can't see how it simplifies to that could you add each step please its killing me
We know that $\overrightarrow{AP}$ and $\overrightarrow{AQ}$ are colinear

$\overrightarrow{AP}$ coordinates are (2p,p²-1)
$\overrightarrow{AQ}$ coordinates are (2q,q²-1)

Therefore (p²-1)/2p = (q²-1)/2q
q(p²-1) = p(q²-1)
p²q-q-pq²+p=0
pq(p-q)+(p-q)=0
(p-q)(pq+1)=0

6. Originally Posted by running-gag
$\overrightarrow{AP}$ coordinates are (2p,p²-1)
$\overrightarrow{AQ}$ coordinates are (2q,q²-1)
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

Originally Posted by running-gag
Therefore (p²-1)/2p = (q²-1)/2q
I'm also unsure how this is derived :/

sorry if this is really simple math think i may need to take a break

7. If $A(x_A,y_A)$ and $B(x_B,y_B)$ then $\overrightarrow{AB}(x_B-x_A,y_B-y_A)$

Here $A(0,1)$ and $P(2p,p^2)$ therefore $\overrightarrow{AP}(2p,p^2-1)$

Now if $\vec{u}(u_x,u_y)$ and $\vec{v}(v_x,v_y)$ are colinear then $u_x\:v_y - u_y\:v_x = 0$

You can also use the slope $\frac{u_y}{u_x} = \frac{v_y}{v_x}$ provided that $u_x \mbox{and} v_x \neq 0$

8. Originally Posted by running-gag
You can also use the slope $\frac{u_y}{u_x} = \frac{v_y}{v_x}$ provided that $u_x \mbox{and} v_x \neq 0$

Is this also only if colinear?

9. Yes

$u_x\:v_y - u_y\:v_x = 0 \Rightarrow u_x\:v_y = u_y\:v_x$ and dividing by $u_x\:v_x$ leads to $\frac{v_y}{v_x} = \frac{u_y}{u_x}$

10. Thanks dude!! real help