parametric questions (a whole bundle!) (1 Viewer)

[mitsui]

first one: (LOLZ! - sorri i am realli havin trouble wif parametrics)

P is given point with parameter p on the parabola x^2=4ay, with focus S. A Line is drawn from S, perpendicular to SP and meets the normal at P in the point Q. PN, QM are drawn perpendicular to the axis of the parabola.
a)find the ordinate at Q
b)If D is the intersection of the directrix and the axis of the parabola, prove that DM=2DN.

thanks guys. =D

 

[Riviet]

Whoah. So many letters, i'll have a go at the first part:

m_{at P}=p
.: m_normal at P=-1/p
let P be the point with parametric coordinates (2ap,ap²
.: equation of normal to P is y-ap²=-1/p(x-2ap)
py-ap³=-x+2ap
x+py=2ap+ap³ (1)

Now coordinates of S are (0,a) ie the coordinates of the focus of x²=4ay
.: m_PS=(ap²-a) / (2ap-0)
=(p²-1)/2p
.: m_{normal to S}=-2p/(p²-1)
.: equation of normal to S is y-a=[-2p/(p²-1)](x-0)
y=-2px/(p²-1) + a (2)

Now you just have to solve (1) and (2) simultaneously, eliminating p is gonna be hard. Where did you get this question from and do you have the answer for it?

 

[mitsui]

hahaha
yea
it is from jones and couchman

...
umm answer to a) is (ap(1-p^2), a(ap^2+1)

 

[Riviet]

Oh good, i thought you had to express Q's coordinates without p in it. Solving the two equations should give you the coordinates then.

 

[Riviet]

Ok lemme see if i get your answer lol:
x+py=2ap+ap³ (1)
y=-2px/(p²-1) + a (2)
rearranging (1),
y=(2ap+ap³-x)/p (3)
Sub (3) into (2),
(2ap+ap³-x)/p=-2px/(p²-1) + a
multiplying everything by p(p²-1),
2ap³-2ap+ap⁵-ap³-p²x+x=-2p²x+ap³-ap
p²x+x=ap-ap⁵
x(p²+1)=ap(1-p⁴)
x=ap(1+p²)(1-p²) / 1+p²
x=ap(1-p²), as required!!

Sub x into (3)
y=[2ap+ap³-ap(1-p²)] / p
Expanding and simplifying we obtain
y=a+2ap²
y=a(2p²+1)
I checked the y coordinate and i'm pretty mine's right, so i think you typed the answer wrong lol.