ext 1 stuff belongs in ext 1 forum
hmmm no wonder no1's attempted this quesiton b4 me
sorry, im lazy. theres heaps of working out so ill just tell u wot i did
x = 2t y = t^2
at a glance theres enough information to find t
get the equation of normal at P
solve simultaneously with x^2 = 4y to find Q (Don't solve quadratic, u have one root as x = 2t, so use the sum or product of roots to find the x co-ordinate of Q)
u get Q(-(4 + 2t^2)/t , ((t^2 + 2)^2)/t^2)
Then use the definition of the parabola (focal length into distance to directrix) to get distances QS and PS
u shud have PS = t^2 +1
QS = ((t^2 + 2)^2)/t^2 + 1 (just the y coordinate + 1)
Use ur QS = 2PS and simplify
u shud get the quartic t^4 - 3t^2 - 4 = 0
Factorise and cancel the factor (t^2 + 1) out (t is real)
u get t = 2 or -2
Sub t = 2 into the points (doesnt matter if t = -2 cos parabola is symmetrical about axis)
U get P(4,4) Q(6,-9) and S(0,1)
Gradient QS * Gradient PS = -1. QED for part a)
Now for part b:
Since u know where P and Q are from part a) its easy sailing from here
just find the tangent equations, solve simultaneously to get T, then distance formula and they are equal
I think there may be an easier way to do it, hopefully the next person isnt as shabby as I am
