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Year 11 Parabola Question (1 Viewer)

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Sep 9, 2005
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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..

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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.
 
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Joined
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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

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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

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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
 
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