1. ## Hard Parametrics Problem

P(2ap, ap^2) and Q(2aq,aq^2) are 2 points on the parabola x^2 = 4ay. Tangents to the parbaola at P and Q intersect at the point T.

a) Show that the equation of the tangent at P is y=px-ap^2.
b) Find the coordinates of T.
c) P and Q move on the parabola so that the line PQ passes through the point (2a,-a). Show that p + q + 1 = pq.
d) Hence, by finding the Cartesian equation of the locus T, show that T lies on a straight line.
e) With the aid of a diagram, carefully explain why the locus of T is not all of the straight line.

Could someone please show me how to do (e). I've found out that the locus of T is x-y+a = 0 and drawn myself a diagram with all the info provided and obtained but I have no clue how to use these to prove (e).

2. Hello xwrathbringerx
Originally Posted by xwrathbringerx
P(2ap, ap^2) and Q(2aq,aq^2) are 2 points on the parabola x^2 = 4ay. Tangents to the parbaola at P and Q intersect at the point T.

a) Show that the equation of the tangent at P is y=px-ap^2.
b) Find the coordinates of T.
c) P and Q move on the parabola so that the line PQ passes through the point (2a,-a). Show that p + q + 1 = pq.
d) Hence, by finding the Cartesian equation of the locus T, show that T lies on a straight line.
e) With the aid of a diagram, carefully explain why the locus of T is not all of the straight line.

Could someone please show me how to do (e). I've found out that the locus of T is x-y+a = 0 and drawn myself a diagram with all the info provided and obtained but I have no clue how to use these to prove (e).
The tangents cannot intersect at a point 'inside' the parabola, hence the locus is those segments of the straight line that lie 'outside' the parabola.

3. How about when P and Q are the same points Grandad?

4. Hello Sunyata
Originally Posted by Sunyata
$\displaystyle (p-q)x=a(p^2-q^2)$
$\displaystyle =a(p-q)(p+q)$
We can divide both sides by $\displaystyle (p-q)$ to get the solution:
$\displaystyle x=a(p+q)$
only if $\displaystyle p\ne q$; otherwise we should be dividing by zero - and that's not allowed!