If the normals at P, Q, R of the parabola `y^2=4ax` meet in O and S be its focus, then prove that `.SP . SQ . SR = a . (SO)^2`.

Explanation

Let the coordinates of A be (h,k).
Equation of normal at `(at^(2),2at)` is
`y=-tx+2at+at^(3)`
If this normal passes through point A, then
`at^(3)+(2a-h)t-k=0` (1)
If `t_(1),t_(2)andt_(3)` are roots of (1), then
`at^(3)+(2a-h)t-k=a(t-t_(1))(t-t_(2))(t-t_(3))` (2)
For these roots, points on the parabola are `P(t_(1)),Q(t_(2))andR(t_(3))`.
Now, put `t=i=sqrt(-1)` in equation (2).
`:." "-ai+(2a-h)i-k=a(i-t_(1))(i-t_(2))(i-t_(3))`
`rArr" "|(a-h)i-k|=a|(i-t_(1))(i-t_(2))i-t_(3))|`
`rArr" "sqrt((a-h)^(2)+k^(2))=asqrt(1+t_(1)^(2))sqrt(1+t_(2)^(2))sqrt(1+t_(3)^(2))`
`rArr" "sqrt(a)sqrt((a-h)^(2)+k^(2))=sqrt(a+at_(1)^(2))sqrt(a+at_(2)^(2))sqrt(a+at_(3)^(2))`
Squaring both sides, we get
`aSA^(2)=SP*SQ*SR`