Here's another solution. I'm just going to make sure you understand what you're doing. I'm a big fan of understanding maths as it will help you A LOT >_>. Another thing to remember is that you should pick the FASTEST method in solving problems. determinisitc has found a very fast and elegant way in solving the problem which is much more favourable.
Now the first thing to understand is why we use simultaneous equations to find the "points of intersections"
In your junior years you learnt that in order to find an x intercept you let y = 0 right? BUT WHY?
If we look further into the maths, by letting y=0 we are actually going through the process of simultaneous equations !
For example, if I gave you the curve y=5x+7 and asked you to find the where it cut the x-axis you'd let y = 0.
But looking deeper we can see that y=0 actually REPRESENTS the x -axis. So we're finding the P.O.I of the X-AXIS and the CURVE.
In other words
y=5x+7......(1)
y=0...........(2)
(1) --> (2)
5x+7=0 (This is another name for "Let y=0")
The thing to learn from this is: - When you use simultaneous equations you are creating a NEW EQUATION which will give you the points of intersections which are REPRESENTED by the roots of the NEW EQUATION.
The new equation we produced was : 5x+7=0 and the only root of this equation was x=-1.4 which represents the P.O.I of the x-axis and the curve, but more formally: the x-intercepts
So fast forwarding we have two curves:
1) x²=4ay [The parabola]
2) x+py=ap³+2ap [The normal at P]
Now by using simultaneous equation (and eliminating y) we get a NEW EQUATION:
(p)x²+(4a)x-(4a²p³+8a²p)=0
If I was to find the X VALUES of this NEW EQUATION it will give me the P.O.I of the two curves.
But we know that x=2ap is a solution to the new equation.
Hence we can use Product of Roots to find the other x value
I.e.
-(8a²p+4a²p³)/p = 2ap.α
α=[-2a(p²+2)]/p
Hence x=[-2a(p²+2)]/p
To find the y coordinate we can subsitute the new x value into any one of the two equations given by the 2 curves. This is because the curves BOTH lie on the same y-coordinate WHEN they're at that SPECIFIC x-coordinate. So any will do.
.'. y = a(p²+ 2)²/p²
.'. Q ([-2a(p²+2)]/p , a(p²+ 2)²/p² )
Ok! So the last step, determining how to represent the parameter Q in terms of the parameter p. Do you understand this?
On the parabola we can place MANY parameters, all of which define the parabola at a certain instance.
It is memorised that the STANDARD FORM of coordinates in terms of parameters is ALWAYS (2aΩ,aΩ²)
So:
-If at a particular point P with parameter p, it will have coordinates (2ap,ap²)
-If at a particular point Q with parameter q, it will have coordinates (2aq,aq²)
-If at a particular point T with parameter t, it will have coordinates (2at,at²)
You get the point right?
So at Q it should have a x coordinate of 2aq.
But we found out that Q is a point of intersection and ALSO has an x coordinate of -2a(p²+2)/p
So we can say that these two x coordinates are equal!
and so:
2aq = -2a(p²+2)/p
We solve for q now so we can express it in terms of the parameter p
q = -(p²+2)/p
That's that!
I hope you understood that. Sorry if I made you feel dumb with the understanding. But the clearer it is of what you're doing, the more likely you are to find tricks and fast methods in doing your problems as deterministic did. Understanding is the key to mathematics.
Good luck!
!