Locus of Parabola (1 Viewer)

[chriscity1]

What did yous get? I got y=-a as R is on the directrix

 

[trickpat23]

I got a strt line but it wasnt that simple but i think i over complicated the question to begin with as nothing seemed to cancelled out for me

 

[Sweet_Lemon]

didn't even prove tat stoopid question.... well did part b~~

 

[The Bograt]

Mine only had 'p's and 'q's in it, so I'm guessing it's wrong....

 

[withoutaface]

mine was in terms of p.

 

[johnamos123]

Yeah answer is y = -a.
cos pq = -1 and R [a(p+q),apq]
use y = apq and sub in pq = -1 you get y = -a

 

[fantasia]

pq=-4

therefore y=-4a

 

[acmilan]

This is how i did it:
PO is perpendicular to QO


gradient of PO = p/2
gradient of QO = q/2


gradient of PO * gradient of QO = -1

p/2*q/2 = - 1
pq = -4

Now at R, y = apq

pq = -4

hence y = -4a

so the locus is y = -4a

I have no idea if thats right

 

[Rorix]

y=-4a is correct.
i don't think you need to mention that p=/=q

 

[johnamos123]

Yeah answer is y = -a.
cos pq = -1 and R [a(p+q),apq]
use y = apq and sub in pq = -1 you get y = -a

nah i was wrong. shit. i done the tangents are at right angles

 

[eska]

wat the hell, pq=-4????


damn i used tangent as the lines gradient????


hell!!! wat am i thinking???

 

[Plebeian]

I did what acmilan did.