1. ## geometry of parabolas

The points P(2ap,ap^2) and Q(2aq, aq^2) vary on the parabola x^2=4ay. The chord PQ subtends a right angle at the vertex. The tangents at P and Q meet at T, while the normals at P and Q meet at N
Find the cartesian equation of the locus of N

So in this qs I have already proved that pq = -4 and that T has coordinates (a(p+q),apq) and that N has coordinates (-apq(p+q),a(p^2+pq+q^2+2))
i just need to use the coordinates of N and find the locus, also using T and pq but i keep getting it wrong, if someone could help me, i would REALLY appreciate it, the answer is x^2 = 16a(y-6a)

thankyou !! =D

2. Originally Posted by flyinhigh123
The points P(2ap,ap^2) and Q(2aq, aq^2) vary on the parabola x^2=4ay. The chord PQ subtends a right angle at the vertex. The tangents at P and Q meet at T, while the normals at P and Q meet at N
Find the cartesian equation of the locus of N

So in this qs I have already proved that pq = -4 and that T has coordinates (a(p+q),apq) and that N has coordinates (-apq(p+q),a(p^2+pq+q^2+2))
i just need to use the coordinates of N and find the locus, also using T and pq but i keep getting it wrong, if someone could help me, i would REALLY appreciate it, the answer is x^2 = 16a(y-6a)

thankyou !! =D
Given

$pq=-4$ (1)

$x=-apq(p+q)$ (2)

$y=a(p^2+pq+q^2+2)$ (3)

we must eliminate $p$ and $q$ to obtain an equation connecting $x$ and $y$.

Subst from (1) into (2):

$x = 4a(p+q)$

$\Rightarrow x^2=16a^2(p+q)^2$

$=16a^2(p^2+2pq+q^2)$

$=16a^2(p^2 +pq +q^2 +2+pq-2)$

$=16a\times \Big(a(p^2+pq+q^2+2) + a(pq-2)\Big)$

$=16a(y -6a)$, from (1) and (3)