MedVision ad

A weird locus question. (1 Viewer)

Tsylana

Member
Joined
May 4, 2008
Messages
80
Gender
Male
HSC
2009
This is from Fitzpatrick 2U

A ladder that is 6m long rests with one end on the horizontal ground, and the other end against a vertical wall. Considering the ground and the wall as the X and Y axes respectively find the midpont of the ladder.

Well... I got to the answer... I had a few problems proving it with different methods though and some of the things I tried to do just plain didnt seem to work so... yeah.. discuss :p. I want to see what you people think... XD.


The answers

x^2 + y^2 = 9

0 < x < 3
0 < y < 3

edit edit...

i actually have a problem doing this question, its just really confusing for me to picture what im actually suppose to do o_O".

A is a point where the crcle x^2 + y^2 = 16 cuts the X axis find the locus of the midpoint of all chords on this circle that contain point A

Q 13 from 10 ( c ) 2u fitzpatrick if anyones wondering
 
Last edited:

3unitz

Member
Joined
Nov 18, 2006
Messages
161
Gender
Undisclosed
HSC
N/A
Tsylana said:
This is from Fitzpatrick 2U

A ladder that is 6m long rests with one end on the horizontal ground, and the other end against a vertical wall. Considering the ground and the wall as the X and Y axes respectively find the midpont of the ladder.

Well... I got to the answer... I had a few problems proving it with different methods though and some of the things I tried to do just plain didnt seem to work so... yeah.. discuss :p. I want to see what you people think... XD.


The answers

x^2 + y^2 = 9

0 < x < 3
0 < y < 3
let x0 denote the horizontal distance from the wall to the base of the ladder, and y0 denote the vertical distance from the ground to the top of the ladder.

length of the ladder is 6m which implies:
x02 + y02 = 62 -----(1)

the midpoint has parameters:
x = x0/2 => x0 = 2x -----(2)
y = y0/2 => y0 = 2y -----(3)

sub (2) and (3) into (1):
(2x)2 + (2y)2 = 62
x2 + y2 = 9
 

3unitz

Member
Joined
Nov 18, 2006
Messages
161
Gender
Undisclosed
HSC
N/A
Tsylana said:
i actually have a problem doing this question, its just really confusing for me to picture what im actually suppose to do o_O".

A is a point where the crcle x^2 + y^2 = 16 cuts the X axis find the locus of the midpoint of all chords on this circle that contain point A

Q 13 from 10 ( c ) 2u fitzpatrick if anyones wondering
let (x0, y0) denote a point which lies on the circle.
i.e. x0^2 + y0^2 = 16 -----(1)

the circle cuts the x axis at A (+4, 0)

midpoint has parameters:
x = (x0 + 4) /2 => x0 = 2x + 4 -----(2)
y = y0 /2 => y0 = 2y -----(3)

sub (2) and (3) into (1):
(2x + 4)^2 + (2y)^2 = 16
 

Tsylana

Member
Joined
May 4, 2008
Messages
80
Gender
Male
HSC
2009
I figured out the first one I was just more or less wondering where the restriction came about o_O"
 

lolokay

Active Member
Joined
Mar 21, 2008
Messages
1,015
Gender
Undisclosed
HSC
2009
3unitz said:
let (x0, y0) denote a point which lies on the circle.
i.e. x0^2 + y0^2 = 16 -----(1)

the circle cuts the x axis at A (+4, 0)

midpoint has parameters:
x = (x0 + 4) /2 => x0 = 2x + 4 -----(2)
y = y0 /2 => y0 = 2y -----(3)

sub (2) and (3) into (1):
(2x + 4)^2 + (2y)^2 = 16
that simplifies to (x + 2)^2 + y^2 = 4
just half the radius of the old circle to find the midpoints, and change x and y in accord with the new centre

Tsylana said:
I figured out the first one I was just more or less wondering where the restriction came about o_O"
this one?
0 < x < 3
0 < y < 3

the y bit means that the since the ladder will only be above the ground, y >= 0. I'm not sure how you would know the x bit though, I would think -3<x<0 would be acceptable too. I don't know if you would need to add in the 3 either, since that's pretty much given by the locus
 
Last edited:

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

Top