When I find the equation of the normals. they both have s and t in them. I couldn't eliminate them. So i can't solve them simultaneously.i will give you some steps for the answer, and then come back in a bit, and if you haven't got it by then i will post the answer/.
1. Find equations of both normals
2. Find intersection of normals by subtracting equations from each other
3. once you have subtracted equations, factorise both sides. Divide by something to get y on its own. this is your value for y.
4. Substitute y into any equation to find x. factorise appropriately. they will be the coordinates of R
1. use the relation st=-2 to eliminate some s and t's from (x,y)
2. rearrange the x variable, and substitute into the y variable to get the equation.
hope this helps. If you cant get it i will post solution a bit later
one normal will have s's in it, one normal will have t's in it. When you solve them simultaneously, your answers will have both s and t's in them. When you solve simultaneously, you are trying to find x and y both in terms of s and t. It is in the second step that you are finding a relationship betweem x,y and a by eliminating the parameters t and s. You can use the relationship st=-2 in this step. Basically wherever you see and st, just get rid of it. i will post in 10 mins if u cant get it
<a href="http://www.codecogs.com/eqnedit.php?latex=x^2=4ay\\ y=\frac{x^2}{4a}\\ y'=\frac{x}{2a}\\ at~P,~y'=t\\ therefore~the~gradient~of~the~normal~is~m=\frac{-1}{t}\\ point~grad~formula\\ y-at^2=\frac{-1}{t}(x-2at)\\ y-at^2=\frac{-x}{t}@plus;2a\\ yt@plus;x=at^3@plus;2at~-(1)\\ similarly,~ys@plus;x=as^3@plus;2as~-(2)\\ (1)-(2)\\ yt-ys=at^3-as^3@plus;2at-2as\\ y(t-s)=a(t-s)(t^2@plus;ts@plus;s^2)@plus;2a(t-s)\\ y=a(t^2@plus;ts@plus;s^2)@plus;2a\\ sub~into~(1)\\ x=at^3@plus;sat-t(a(t^2@plus;ts@plus;s^2)@plus;2a)\\ =at^2s-ats^2\\ R(ats(t-s),a(t^2@plus;ts@plus;s^2)@plus;2a)\\ \\ we~know~ts=-2\\ therefore~R(-2a(t-s),a(t^2@plus;s^2-2)@plus;2a)\\ taking~x=-2a(t-s)\\ \frac{x}{-2a}=(t-s)\\ squaring~both~sides~\frac{x^2}{4a}=t^2@plus;4@plus;s^2\\ substituting~into~y\\ y=a(\frac{x^2}{4a}-6)@plus;2a\\ =\frac{x^2}{4}-4a\\ therefore~the~locus~of~R~is\\ x^2=4(y@plus;4a)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x^2=4ay\\ y=\frac{x^2}{4a}\\ y'=\frac{x}{2a}\\ at~P,~y'=t\\ therefore~the~gradient~of~the~normal~is~m=\frac{-1}{t}\\ point~grad~formula\\ y-at^2=\frac{-1}{t}(x-2at)\\ y-at^2=\frac{-x}{t}+2a\\ yt+x=at^3+2at~-(1)\\ similarly,~ys+x=as^3+2as~-(2)\\ (1)-(2)\\ yt-ys=at^3-as^3+2at-2as\\ y(t-s)=a(t-s)(t^2+ts+s^2)+2a(t-s)\\ y=a(t^2+ts+s^2)+2a\\ sub~into~(1)\\ x=at^3+sat-t(a(t^2+ts+s^2)+2a)\\ =at^2s-ats^2\\ R(ats(t-s),a(t^2+ts+s^2)+2a)\\ \\ we~know~ts=-2\\ therefore~R(-2a(t-s),a(t^2+s^2-2)+2a)\\ taking~x=-2a(t-s)\\ \frac{x}{-2a}=(t-s)\\ squaring~both~sides~\frac{x^2}{4a}=t^2+4+s^2\\ substituting~into~y\\ y=a(\frac{x^2}{4a}-6)+2a\\ =\frac{x^2}{4}-4a\\ therefore~the~locus~of~R~is\\ x^2=4(y+4a)" title="x^2=4ay\\ y=\frac{x^2}{4a}\\ y'=\frac{x}{2a}\\ at~P,~y'=t\\ therefore~the~gradient~of~the~normal~is~m=\frac{-1}{t}\\ point~grad~formula\\ y-at^2=\frac{-1}{t}(x-2at)\\ y-at^2=\frac{-x}{t}+2a\\ yt+x=at^3+2at~-(1)\\ similarly,~ys+x=as^3+2as~-(2)\\ (1)-(2)\\ yt-ys=at^3-as^3+2at-2as\\ y(t-s)=a(t-s)(t^2+ts+s^2)+2a(t-s)\\ y=a(t^2+ts+s^2)+2a\\ sub~into~(1)\\ x=at^3+sat-t(a(t^2+ts+s^2)+2a)\\ =at^2s-ats^2\\ R(ats(t-s),a(t^2+ts+s^2)+2a)\\ \\ we~know~ts=-2\\ therefore~R(-2a(t-s),a(t^2+s^2-2)+2a)\\ taking~x=-2a(t-s)\\ \frac{x}{-2a}=(t-s)\\ squaring~both~sides~\frac{x^2}{4a}=t^2+4+s^2\\ substituting~into~y\\ y=a(\frac{x^2}{4a}-6)+2a\\ =\frac{x^2}{4}-4a\\ therefore~the~locus~of~R~is\\ x^2=4(y+4a)" /></a>
its a bit cluttered lol, hope that clears things up, and i hope its the right answer!
no worries