Locus problems relating to parabola. (1 Viewer)

[frmldhyd]

Hey,

I need help with this question, fsr this topic is just not going my way >_>

"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 mid-point of PQ?"

 

[blackops23]

P(2ap, ap^2)
Q(2aq, aq^2)

Midpoint of PQ = (a(p+q) , a(p^2 + q^2)/2 )
Now condition is: 2ap - 2aq = 2a
therefore: p - q = 1

From parametric coordinates of M
x=a(p+q)
(p+q) = x/a

y = (a(p^2 + q^2)/2
2y/a = p^2 + q^2

therefore:

x^2/a^2 = (p+q)^2 = p^2 + 2pq + q^2
BUT... (p-q)^2 = p^2 - 2pq + q^2
Add these 2 equations together:

(p+q)^2 + (p-q)^2
= 2(p^2 + q^2)
=2(2y/a)
= 4y/a

THEREFORE:

(x^2/a^2) - (1)^2 = 4y/a
x^2 + a^2 = 4ay

Cheers

 

[frmldhyd]

thank you so much...

you're a genius, i think you have cleared up something for me!

greatly appreciated.

 

[Bieber]

Excuse me @blackops23/blackops23, I don't get where you got the (x^2/a^2) - (1)^2 = 4y/a from? (@frmldhyd/frmldhyd can help too?)

 

[Aesytic]

Excuse me @blackops23/blackops23, I don't get where you got the (x^2/a^2) - (1)^2 = 4y/a from? (@frmldhyd/frmldhyd can help too?)

it is meant to be (x^2/a^2) + (1)^2 = 4y/a

this is from the line before where he had (p+q)^2 + (p-q)^2 = 4y/a
p+q = x/a, so (p+q)^2 = x^2/a^2
and from the condition, p-q = 1, so (p-q)^2 = 1^2
substituting those back in, you get the equation (x^2/a^2) + (1)^2 = 4y/a