parametrics (1 Viewer)

Mu5hi

Banned
Joined
Apr 15, 2009
Messages
425
Location
sydney
Gender
Male
HSC
2010
Book 2 Couchman and Jones
Exercise 23.2
Q2

Can anyone willing to help me, i cant remember how to do it :(
 

Mu5hi

Banned
Joined
Apr 15, 2009
Messages
425
Location
sydney
Gender
Male
HSC
2010
find the equation of the chord joining the points t(1)=2 and t(2)= -1/2 on the porabola x^2=y

simple yes, but i ahvent been taught this. so yeah
 

randomnessss

Member
Joined
Oct 11, 2008
Messages
91
Gender
Male
HSC
2010
Uni Grad
2015
The general form of a parabola with vertex (0,0) is x2 = 4ay. Therefore:
4a = 1
a = 0.25

For the parametric equations, sub a = 0.25:
x = 2at
x = 0.5t - (A)

y = at^2
y = 0.25t^2 - (B)

Sub t(1) in (A) and (B). This will form an x-ordinate and y-ordinate (i.e. a set of co-ordinates)


Sub t(2) in (A) and (B). This will form an x-ordinate and y-ordinate (i.e. a set of co-ordinates)

Use the gradient formula treating the x-ordinate and y-ordinate of each t value as a point.
 

addikaye03

The A-Team
Joined
Nov 16, 2006
Messages
1,267
Location
Albury-Wodonga, NSW
Gender
Male
HSC
2008
find the equation of the chord joining the points t(1)=2 and t(2)= -1/2 on the porabola x^2=y

simple yes, but i ahvent been taught this. so yeah
above or:

Set up a table so that we got t(1)=2, x=2at where a=0.25 =1/2(t)=1 then y= 1^2=1

For t(2)=-1/2, x=2at so =1/2(t)=-1/4 then y= 1/16

So now you got (x2,y2) and (x1,y1)

Now just use (y-y1)/(x-x1)=(y2-y1)/(x2-x1) [2 point form]

EDIT: LOL, didn't look properly at the above method, just realised it's pretty much the same thing haha
 

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

Top