parabola and locus q (1 Viewer)

[ToO LaZy ^*]

Two points P(2ap,ap²) and Q(2aq,aq²) lie on the parabola x² = 4ay, where A>0. The chord PQ passes through the focus.

(c) show that the chord PQ has length A(p+1/p)²

 

[CrashOveride]

Distance formula. pq=-1 then q=-1/p

Factorise and think about (p+1/p)^4 expansion using binomial thm.

 

[VieTz_88]

Hhehe this is how I did it
Perpendicular distance from point Q(2aq,aq2) to directrix is equal to distance from from the focus(0,a) to point Q , ie. distance = aq2 + a
Similarly, perpendicual distance from point P(2ap,ap2) to directrix(y=-a) is : ap2+a .:. the distance of the chord PQ = distance Q + distance P from directrix = ap2+aq2+2a
=a(p2+q2+2)


gradient of PQ = ap2-aq2/2ap-2aq= a(p-q)(p+q)/2a(p-q)= p+q/2
equation of PQ : y-ap2 = (p+q/2)(x-2ap)
y = x(p+q)/2 -ap2+apq+ap2
.:.y = x(p+q)/2 +apq

As focus(0,a) is a point on the chord PQ, sub (0,a) into y=x(p+q)/2+apq
we get pq=-1
.:.q=-1/p
q2=1/p2
sub this into distance of chord PQ : a(p2+q2+2)
=a(p2+1/p2+2)
=a(p+1/p)2

 

[CrashOveride]

pq is actually -1.

And use some tags or something to facilitate coherency ^^

 

[VieTz_88]

ahhh thanks crash for that
umm what are tags im kinda new to this hehe

 

[ToO LaZy ^*]

ah.got it..thanks people.

 

[CrashOveride]

for powers: go <#sup> to begin and to finish </#sup> (Note: dont use the "#" i put it there so it would show up).

For subscript: replace sup with sub

 

[ToO LaZy ^*]

you need to use < / sup> to finish. (without the spaces)