3:34...............................
<span>Answer:
Its too long to write here, so I will just state what I did.
I let P=(2ap,ap^2) and Q=(2aq,aq^2)
But x-coordinates of P and Q differ by (2a)
So P=(2ap,ap^2) BUT Q=(2ap - 2a, aq^2)
So Q=(2a(p-1), aq^2)
which means, 2aq = 2a(p-1)
therefore, q=p-1
then I subbed that value of q in aq^2
so Q=(2a(p-1), a(p-1)^2)
and P=(2ap,ap^2)
Using these two values, I found the midpoint which was:
M=( a(2p-1), [a(2p^2 - 2p + 1)]/2 )
then x = a(2p-1)
rearranging to make p the subject
p= (x+a)/2a</span>
(-2,3) (2,1) -2-1=-3 3-2=1 so that means mx+-3
<span>The answers are either:
x=<span>2+<span><span><span>√15</span><span> or </span></span>x</span></span></span>=<span>2−<span>√<span>15
Hope this helps!</span></span></span>
