Locus Problem... (1 Viewer)

[~Theta~]

Argh, I've tried to answer this question twice but when I reach the end, it's not correct! So if you could provide a solution, it'll be greatly appreciated.

Q: Find the equation of the locus of point R that is the intersection of the normals at P(2p,p^2) and Q(2q,q^2) on the parabola x^2=4y, given that pq = -4

Man, I should be able to do this, it doesn't seem that complicated

Thanks in advance

 

[Trebla]

Argh, I've tried to answer this question twice but when I reach the end, it's not correct! So if you could provide a solution, it'll be greatly appreciated.

Q: Find the equation of the locus of point R that is the intersection of the normals at P(2p,p^2) and Q(2q,q^2) on the parabola x^2=4y, given that pq = -4

Man, I should be able to do this, it doesn't seem that complicated

Thanks in advance

Equation of normal at P (derive this):
x + py = p³ + 2p
Similarly equation of normal at Q:
x + qy = q³ + 2q
Subtracting the two equations gives:
y(p - q) = p³ - q³ + 2(p - q)
y(p - q) = (p - q)(p² + pq + q²) + 2(p - q)
Assuming p =/= q
y = p² + q² - 4 + 2
= p² + q² - 2
Sub back into any equation
x + p³ + pq² - 2p = p³ + 2p
x + (pq)q = 4p
x = 4(p + q)

So we have:
x = 4(p + q) => p + q = x/4
y = p² + q² - 2 => p² + q² = y + 2
Note that (p + q)² = p² + q² + 2pq
So subbing x and y gives:
(x/4)² = y + 2 - 8
y = x²/16 + 6

 

[~Theta~]

Thanks so much Trebla!