Prove that the length of the intercept on the normal at the point `P(a t^2,2a t)` of the parabola `y^2=4a x` made by the circle described on the line joining the focus and `P` as diameter is `asqrt(1+t^2)` .

Explanation

Equation of normal at `P(at^(2),2at)` to the parabola `y^(2)=4ax` is
`y=-tx+2at+at^(3)` (1)
Let this normal meet the circle described on SP as diameter in N, then `angleSNP=90^(@)` (angle in a semicircle).
`:." "PN^(2)=SP^(2)-SN^(2)`
Now `SP=a+at^(2)`
and SN = Distance of focus from the normal
`(|at-2at-at^(3)|)/(sqrt(1+t^(2)))=atsqrt(1+t^(2))`
`:." "PN^(2)=a^(2)(1+t^(2))^(2)-a^(2)t^(2)(1+t^(2))
=a^(2)(l+t^(2))[l+t^(2)-t^(2)]=a^(2)(l+t^(2))`
`:." "PN=asqrt(l+t^(2))`