parametric locus

[osak23]

hi i have 2 problems for anyone who has time

1. Tangents to the parabola x^2=4ay drawn from points P(2ap,ap^2) and Q(2aq,aq^2) intersect at right angles at point R. Find the locus of
a) point R
b) The mid point of PQ

2.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) in the parabola x=4Y,given that PQ=-4.

the '^2' only applies to p of the y-coord ap^2.
the '^2' only applies to q of the y-coord aq^2.

[Timske]

Q1.
Find the point of intersection of tangents at P and Q to find the coordinates of R.
P(2ap, ap^2) and Q(2aq, aq^2)

x^2 = 4ay
y = x^2/4a
y' = x/2a
Sub x values in x/2a to to find gradient
2ap/2a = p , 2aq/2a = q
Gradient of tangent at P = p
Gradient of tangent a Q = q

using point gradient formula , y-y1 = m(x - x1)
Tp; y - ap^2 = p(x - 2ap)
Tp = px - ap^2
therefore, Tq = qx - aq^2

solve simulataneously
y = px - ap^2 ...1
y = qx - aq^2 ...2
1 - 2
- px + qx + ap^2 - aq^ 2 = 0
a(p^2-q^2) = px - qx
[a(p-q)(p+q)/(p-q)] = x
x = a(p+q)
SUB INTO 1
y - pa(p+q) + ap^2 = 0
y - ap^2 - apq + ap^2 = 0
y = apq
R is (a(p+q), apq)
Locus, y = -4a



2. Midpoint of PQ use midpoint formula [(x1 + x2)/2 , (y1 + y2)/2]
P(2ap,ap^2) and Q(2aq,aq^2)

[(2ap + 2aq)/2 , (ap^2+aq^2)/2]
M - [ap+aq , (ap^2 + aq^2)/2]

[Nooblet94]

Q1.
Find the point of intersection of tangents at P and Q to find the coordinates of R.
P(2ap, ap^2) and Q(2aq, aq^2)

x^2 = 4ay
y = x^2/4a
y' = x/2a
Sub x values in x/2a to to find gradient
2ap/2a = p , 2aq/2a = q
Gradient of tangent at P = p
Gradient of tangent a Q = q

using point gradient formula , y-y1 = m(x - x1)
Tp; y - ap^2 = p(x - 2ap)
Tp = px - ap^2
therefore, Tq = qx - aq^2

solve simulataneously
y = px - ap^2 ...1
y = qx - aq^2 ...2
1 - 2
- px + qx + ap^2 - aq^ 2 = 0
a(p^2-q^2) = px - qx
[a(p-q)(p+q)/(p-q)] = x
x = a(p+q)
SUB INTO 1
y - pa(p+q) + ap^2 = 0
y - ap^2 - apq + ap^2 = 0
y = apq
R is (a(p+q), apq)
Locus, y = -4a



2. Midpoint of PQ use midpoint formula [(x1 + x2)/2 , (y1 + y2)/2]
P(2ap,ap^2) and Q(2aq,aq^2)

[(2ap + 2aq)/2 , (ap^2+aq^2)/2]
M - [ap+aq , (ap^2 + aq^2)/2]

Locus of the first one should be y=-a, shouldn't it?

[Timske]

Uh yeahy bad because pq=-1

[osak23]

hey how did you establish pq=-1?

[Aesytic]

because the tangents are perpendicular, then the gradients of the tangents must be equal to -1
since the gradients of the tangents to the curve at P and Q are p and q respectively, therefore pq=-1

[osak23]

ooooooo lol .
thanks