• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

parametric question from 2007 trial (1 Viewer)

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
P & Q are points on the parabola x^2 = 4ay with parameter value t= 1 and t=2. Show that the normals to the parabola Q and P intersect at a point R on the parabola.

Thanks in advance.
 

3unitz

Member
Joined
Nov 18, 2006
Messages
161
Gender
Undisclosed
HSC
N/A
lyounamu said:
P & Q are points on the parabola x^2 = 4ay with parameter value t= 1 and t=2. Show that the normals to the parabola Q and P intersect at a point R on the parabola.

Thanks in advance.
dy/dx = x/2a
mnormal = -2a/x
y - y0 = -2a/x0 . (x - x0)

normal at P (2a, a):
y - a = -2a/2a (x - 2a)
y - a = -x + 2a
y = -x + 3a ----------(1)

normal at Q (4a, 4a):
y - 4a = -2a/4a (x - 4a)
y - 4a = -x/2 + 2a
y = - x/2 + 6a ----------(2)

equate (1) and (2):
-x + 3a = -x/2 + 6a
-x/2 = 3a
x = -6a
sub into (1) to get y:
y = 9a

.'. R = (-6a, 9a)

check if lies on parabola x^2 = 4ay:
(-6a)^2 =? 4a(9a)
36a^2 =? 36a^2
true, therefore normals intersect at R which lies on the parabola.
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
3unitz said:
dy/dx = x/2a
mnormal = -2a/x
y - y0 = -2a/x0 . (x - x0)

normal at P (2a, a):
y - a = -2a/2a (x - 2a)
y - a = -x + 2a
y = -x + 3a ----------(1)

normal at Q (4a, 4a):
y - 4a = -2a/4a (x - 4a)
y - 4a = -x/2 + 2a
y = - x/2 + 6a ----------(2)

equate (1) and (2):
-x + 3a = -x/2 + 6a
-x/2 = 3a
x = -6a
sub into (1) to get y:
y = 9a

.'. R = (-6a, 9a)

check if lies on parabola x^2 = 4ay:
(-6a)^2 =? 4a(9a)
36a^2 =? 36a^2
true, therefore normals intersect at R which lies on the parabola.
Thank you so much!
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
A particle is moving in a straight line. Initially the particle is at a fixed point O on the line. At time t seconds, it has displacement x metres from O, velocity vm/s given by v = 10-x and acceleration a m/s^2

(1) find a in terms of x = finished
(2) use integration to show that x = 10-10e^(-t) = finished
(3) find the limiting position of the particle and the time it takes to move within 1 cm of this limiting position.

I only need help for (3). I put (1) and (2) as a reference.

I don't understand what it means by "limiting" position.

Thanks in advance.
 

Aerath

Retired
Joined
May 10, 2007
Messages
10,169
Gender
Undisclosed
HSC
N/A
Like, he's done it, and he doesn't need help with those questions.
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
u-borat said:
wat does it mean by 'finished' in parts 1) and 2)?
I already did those questions. They were worth 1 mark and 2 marks respectively. I think the question I am asking is one mark but I want to know how to do this. Anyone?
 

u-borat

Banned
Joined
Nov 3, 2007
Messages
1,755
Location
Sydney
Gender
Male
HSC
2013
lol my bad, i thought they were part of the question.

with the question,

as t---> infinity, e^-t-->0.
therefore, x approaches 10 from below.

then, just sub x=10.01 and solve for t.
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
u-borat said:
lol my bad, i thought they were part of the question.

with the question,

as t---> infinity, e^-t-->0.
therefore, x approaches 10 from below.

then, just sub x=10.01 and solve for t.
sweet man, I will try that. :eek:
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top