# Thread: Parametric problems and the Parabola

1. ## Parametric problems and the Parabola

I've got a one mark question on an assignment, so 'logic' dictates that it should be simple, right? But I just don't get it.

I've got the variable point P(2at, at^2), which lies on the parabola x^2=4at

I've already proven that the equation of the normal at P is x+ty=at^3+2at

Now, based on that, I have to find the co-ordinates of Q (also on the parabola), whose normal is perpendicular to the normal at P.

The next part of the assignment gives me the point at which the two normals intersect (R (a[t-1/t] , a[t^2+1+1/t^2]), so I know that the equation of QR is y=tx+a/t^2+a, but I seeing as that is backwards-engineering from a 4 mark question, there's got to be something I'm missing in the "find Q" one.

2. You don't have to do all that.

Equation of the normal to the parabola$\displaystyle x^2=4ay$

for the parameter $\displaystyle t$ is $\displaystyle x+ty=at^3+2at$

Now the slope of the normal is =$\displaystyle -1/t$

So for any parametric point "t" on the given parabola,the slope of the normal to the parabola at that point is -1/t

Now we need the point Q having parameter t'.

Now as the normal at Q is perpendicular to that at P, (-1/t')(-1/t)=-1

So, t'=-1/t

so the coordinates of Q are merely $\displaystyle (2at',a(t')^2)$

where $\displaystyle t'=-1/t$