MedVision ad

Year 11 Parabola Question (1 Viewer)

Joined
Sep 9, 2005
Messages
195
Location
Kenso
Gender
Male
HSC
2006
hey guys, just strugglin with this one

the question reads

"Find the equation of the locus of point P that moves so that PA is perpendicular to PB where A=(-1,-3) and B=(2,4)"

Alrighty so far i know that (-1,4) and (2,-3) lie on the locus, i think you get the gradients with the two point formula, but im kinda stuck and the question is really starting to shit me so If anyone could help it would be greatly appreciated.

Cheers.
 

Stan..

Member
Joined
Nov 4, 2004
Messages
278
Gender
Male
HSC
2006
Use the 'two point gradient formula', m1m2 = -1. Then a little re-arrangement and you find a solution, I am unsure if it is right. But I got a answer.

Forget that, Trev's method is correct. I considered that before I attempted the problem, Posted a bad method. Sorry.
 
Last edited:
Joined
Sep 9, 2005
Messages
195
Location
Kenso
Gender
Male
HSC
2006
Stan.. said:
Use the 'two point gradient formula', m1m2 = -1. Then a little re-arrangement and you find a solution, I am unsure if it is right. But I got a answer.

Forget that, Trev's method is correct. I considered that before I attempted the problem, Posted a bad method. Sorry.
ive tried that...and i get all this x and y mumbo jumbo that doesnt really make sense.....do u think u guys could be a tad more specific.....thanks.
 

apak

New Member
Joined
Oct 19, 2005
Messages
25
Gender
Male
HSC
2007
Stan.. said:
Use the 'two point gradient formula', m1m2 = -1. Then a little re-arrangement and you find a solution, I am unsure if it is right. But I got a answer.

Forget that, Trev's method is correct. I considered that before I attempted the problem, Posted a bad method. Sorry.
I dont think you use 2 point formula. it doesnt really make sense. the 2 point formula will just give you the gradient of the line joining the to points but not the locus of the point.
 

apak

New Member
Joined
Oct 19, 2005
Messages
25
Gender
Male
HSC
2007
A=(-1,-3) B=(2,4) P=(x,y)
then m of AP = (y+3)/(x+1)
and m of BP = (y-4)/(x-2)
perpendicular at -1 therefore
m of AP x m of BP = -1
[(y+3)/(x+1)][(y-4)/(x-2)]=-1
y²-4y+3y-12/x²-2x+x-2=-1
-x²+x+2=y²-y-12
x²+y²-x-y=14
 
Last edited:

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

Top