Parabola locus (1 Viewer)

[enigma_1]

How do you do 18a and c?
And then for 19a?

 

[rumbleroar]

18a. Use the fact that PS = perpendicular distance of P to the directrix.
18c. If PQ is focal chord, pq=-1. Find PQ (in terms of p and q), then let q=-1/p and substitute.

19a. find the gradient of PS and QS and show they multiply to give -1 (you can find coords of Q by finding out where the normal at P intersects the parabola)

 

[enigma_1]

Thanks, can I get worked solutions? I did that but it didnt' come

 

[MrBeefJerky]

18 a) Equation of parabola: x^2=4y where a= 1
PS = PM where M is a point at the directrix (y=-1) directly below P (same x-coordinate)
PM = p^2-(-1) = p^2+1 = PS
b) PS = p^2+1 and QS = q^2+1
PS + QS = p^2+q^2+2
c) Since PQ is a focal chord pq=-1 (can prove by finding eqn of PQ and substitute the focus (0,1))
distance PQ = p^2+q^2+2
Since pq=-1
q= -1/p
Substitute into distance PQ = p^2+(-1/p)^2+2
= (p+1/p)^2

 

[rumbleroar]

Thanks, can I get worked solutions? I did that but it didnt' come

for which one?