locus q?!? (1 Viewer)

[Trev]

hey, how do u go about doin this?...
Two points P, Q move on the parabola x^2 = 4ay so that the x coordinates of P and Q differ by a constant value 2a. What is the locus of M, the midpoint of PQ?
............hmm

 

[Xayma]

P(2at₁,at₁²)
Q(2at₂,at₂²)

2at₂=2at₁+2a
=2a(t₁+1)
∴ t₂=t₁+1

∴ at₂²=a(t₁²+2t₁+1)

M(x,y)=([2at₁+2a{t₁+1}]/2,[at₁²+a{t₁²+2t₁+1}]/2)
=(a[2t₁+1], a[t₁²+t₁+1/2])

 

[Trev]

ohk, i understand what u did - but how do u find the locus of M?
you wrote it as a point rather than a line?!?
ps. the answer is x^2 = a(4y - a) .......help still plz?

 

[Xayma]

Its a variable point. Didn't you cover (2at, at²) to cover a variable point on the parabola x²=4ay?

But to show you.

(a[2t+1])²=a²[4t²+4t+1]
=a(a[4{t²+t+1/2}-1]
=a(4y-a)

Therefore the equation of the line (without the parameter t) is
x²=a(4y-a)

 

[Trev]

hmm another q!
At a point P on the parabola x^2=4ay, a normal PK is drawn. From the vertex O a perpendicular OM is drawn to meet the normal at M. Show that the equation of the locus of M as P varies on the parabola is: x^4 - 2ax^2y + x^2y^2 - ay^3 = 0