Question

The normal `(2ap, ap^2)` to the parabola `x^2=4ay` meets the curve again at `Q(2aq, aq^2)`?

(a) Show that `q=-(2+p^2)/p`

(b) Find the coordinates of P so that the lines `OP` and `OQ` meet at right angles, where `O` is the origin.

Answer 1

(a)Given equation of the parabola
`x^2=4ay`

`=>(dy)/(dx)=1/(4a)*2x=x/(2a)`

Slope of the normal

`m=-1/[(dy)/(dx)]_(2ap,ap^2)=-(2a)/(2ap)=-1/p`

Again the slope of the normal as it joins `(2ap,ap^2) and (2aq,aq^2)` is also
`m=(ap^2-aq^2)/(2ap-2aq)=1/2(p+q)`

So we have

`1/2(p+q)=-1/p`

`=>q=-2/p-p=-(2+p^2)/p....(1)`

OP and OQ being orthogonal the produt of their slopes `(ap^2)/(2ap)*(aq^2)/(2aq)=-1`

`=>pq=-4....(2)`

From (1) and (2) we get

`-4=-(2+p^2)`

`=>p=sqrt2`

(b)So coordinates of the point P will be `(2sqrt2a,2a)`