Parametric help (2 Viewers)

[MATHSCAPE]

hi, i need help with 2 questions about parametrics. i think they're easy but i can't seem to do them.

1) P(2ap,ap^2) and Q(2aq,aq^2) are two points on the parabola x^2=4ay. Given that the gradient of the chord PQ is a constant m, show that the locus of the mid point of PQ is a line parallel to the y axis.

2) PQ is a chord of x^2=4ay such that the abscissa of P is twice that of Q. Show that the locus of M, the mid point of PQ is 5x^2=18ay.

thanks!

 

[life92]

1. gradient of PQ = (ap^2 - aq^2) / (2ap - 2aq)
= [ a (p-q)(p+q) ] / [ 2a (p-q) ]
= p+q / 2
= m

that is, (p+q)/2 = m

now the midpoint of PQ,

x = (2aq + 2aq) / 2
= a (p+q)

y = (ap^2 + aq^2) / 2
= a (p^2+q^2) / 2

Now, x = a (p+q), but p+q = 2m
therefore, x = 2am, which is a constant

and hence the locus of the midpoint of PQ is a vertical line, x = 2am

2. lol okay
from above
midpoint of PQ
x = a (p+q)
y = a (p^2 +q^2) / 2

since the abscissa of P is 2 times the abscissa of Q,
then if P = (2ap,ap^2) and Q = (2aq, aq^2)
then 2ap = 2 x 2aq
therefore, p = 2q

now,
p+q = x/a
3q = x / a

y = a/2 [ p^2 + q^2 ]
= a/2 [ (p+q)^2 - 2pq ]
= a/2 [ (3q)^2 - 4q^2 ]
= a/2 [ (x/a)^2 - 4 . (x/3a)^2 ]
= a/2 [ x^2 / a^2 - 4x^2 / 9a^2 ]
= a/2 [ 5x^2 / 9a^2 ]
= 5x^2 / 18a

therefore 5x^2 = 18ay

ps. sorry still dno how to use latex XD ,,

 

[MATHSCAPE]

no worries, i understood it. thanks for the help. +rep

 

[Supergirl22]

Hi, I was wondering if anyone could help me with these questions... i'd really appreciate any help asap thanx

The line L passes through the focus of x^2=4ay and makes an angle of 45 with the x axis.
(a) Find the equation of L.
(b) If P is the point (2ap, ap^2) on the parabola and Q is a point on L such that PQ is parallel to the y axis, find the coordinates of Q.
(c) Find the coordinates of M, the midpoint of PQ.
(d) Find the locus of M.

 

[TheMathStudent]

Hope this helps.

Cheers

 

[Supergirl22]

Thanks a lot! You saved me

 

[Supergirl22]

Sorry I have one last question:
If the chord of contact of the tangents to the parabola x^2=4ay from an external point T(x1,y1) meets the directrix at R, prove that RT subtends a right angle at the focus.

 

[TheMathStudent]

Its been a while since I did HSC so what exactly does subtend mean? Is it asking to prove that RTA (A being the focus) is 90 degrees? Also are you sure the question is correct?

Anyway I attempted to answer the question, and this is how you should approach a parametric Q in general but I couldn't complete it because I think there is some information missing.

This should give you some idea of how to do it. Sorry I cant be of any more help.

 

[Supergirl22]

Thanks for the help. Appreciate it