some help with locus please.... (1 Viewer)

[dumarab]

Find the equation of the locus of point R that is the intersection of the normals at P(2p,p²) and Q(2q,q²) on the parabola x²=4y, given that pq = -4



answer at back is
x² = 16(y-6)


Much help is appreciated please.

i just get stuck, please help

 

[Affinity]

gradient of tangent at P is p and that at Q is q, so the gradients of the normals are
-1/p and -1/q
so the equations of the normals are
(x - 2p)(-1/p) = (y - p^2)
(x - 2q)(-1/q) = (y - q^2)

Then you simplify a little

x - 2p = p^3 - py
x - 2q = q^3 - qy

and if you scale them,

qx - 2pq = qp^3 - pqy
px - 2pq = pq^3 - pqy

from the first set of equations (subtracting one from another)

-2(p - q) = (p^3 - q^3) - (p - q)y

dividing by (p-q)
-2 = p^2 + pq + q^2 - y
y = p^2 + pq + q^2 + 2
but pq = -4 so
y = p^2 -2 + q^2

from the second set of equations
(p - q)x = -pq(p^2 - q^2)
x = -pq(p + q)
but pq = -4 so
x = 4(p+q)

now (p+q)^2 = p^2 + q^2 + 2pq = p^2 + q^2 -8

so (x/4)^2 = y - 6
so x^2 = 16(y - 6)

The procedure here is standard and should be straightforward

 

[dumarab]

The normal at any point P(-8p,4p²) on the parabola x²=-16y cuts the y-axis at point M. find the equation of the locus of the midpoint of PM

I got another one, but his time i think the answers in the book are rong because i'm so close

the answer is

x² =-4(y+4)

 

[tommykins]

回复: Re: some help with locus please....

No, the answer in the book is correct.

Heres the working out -