Chord of tangent - parametric help needed (1 Viewer)

[Petinga]

1. The chord of contact to the parabola x^2=4ay from the point P(x subscript zero, y subscript o) passese through the point Q (0,2a). Show that the locus of the mid point of PQ is the x axis.

2. Show that the chord of contact of tangents from the point (a,-a) to the parabola x^2=4ay has length 5a.

3. P and Q are points on the parabola x^2=4ay. Tangents TP and TQ are drawn from an external point T and these tangents cut the x-axis at A and B. Show that the line joining the focus to the midpoint of AB is perpendicuklar to the chord of contact PQ.

 

[Angelo0523]

LoL....they are just the last three questions from the 3u fitzpatric book

 

[Mountain.Dew]

okay, here goes...

1) realise that the chord of contact at P is x x0 = 2a (y - y0). it passes through Q(0,2a). so, SUB, x=0, y=2a. --> (0)(x0) = 2a(2a-y0)

so 0 = 2a(2a-y0)
0=2a-y0
y0=2a --> REALISE that both P and Q have the same y-value + IT IS A CONSTANT.

So, any pt along PQ will be restricted to the line y=2a, HENCE the x-axis.

(have my doubts about this reasoning, please correct if necessary)

2) this one is a little bit tricky, this is prolly the long way of doing it, but it works. we have any two general pts, P(2ap,ap^2) and Q(2aq,aq^2).

find the intersection of their tangents. we get x=a(p+q) and y=apq.

we know that x=a, (from the question) so a=a(p+q), p+q = 1....(A)
we know that y=-a (question), so (-a) = a(pq), pq = -1......(B)

now, use distance formula

(distance)^2 = (2ap - 2aq)^2 + (ap^2 - aq^2)^2
(distance)^2 = 4a^2(p-q)^2 + a^2((p+q)(p-q))^2 [perfect square]
(distance)^2 = 4a^2(p-q)^2 + a^2((1)(p-q))^2 [using (A)]
(distance)^2 = (p-q)^2 (4a^2 + a^2) [factorising the (p-q)^2 thing out]
knowing that (p-q)^2 = (p+q)^2 - 4pq, = 1 - 4(-1) = 5 (using (A) and (B))
(distance)^2 = 5(5a^2)
(distance)^2 = 25a^2

hence, distance/length of chord of contact of tangets = 5a.

i'll post this up first, then q3 will come shortly in next post

 

[Mountain.Dew]

3)

okay, we have pt P(2ap,ap^2) and Q(2aq,aq^2)

gradient PQ = (2ap - 2aq) / (ap^2 - aq^2)
gradient PQ = 2a(p-q) / a(p+q)(p-q)
gradient PQ = 2/(p+q)

we have this handy, so that we can show two gradients times together = -1

now, tangent at P is: y=px-ap^2. cuts at A, where y=0 SO sub
--> get 0 = px-ap^2 --> x=ap SO A = (ap,0)
similarily, at B, B = (aq,0)

denote midpt AB as M.
M = a(p+q)/2,0
S= (0,a)
now, gradient SM = (a(p+q)/2 - 0)/(0-a)
so gradient SM = -(1/2)(p+q)

THEREFORE: gradient PQ * gradient SM = 2/(p+q) * -(1/2)(p+q)
and lucky for us, gradient PQ * gradient SM = -1
HENCE line joining the focus to the midpoint of AB is perpendicular to the chord of contact PQ. (just quoting the question LOL )