Find the locus of the midpoint of normal chord of parabola `y^2=4a xdot`

Explanation

Normal to parabola `y^(2)=4ax` at point P(t) meets the parabola again at point Q(t') such that `t'=-t-(2)/(t)` (1)
Let midpoint of PQ be R (h,k).
`:." "h=(at^(2)+at'^(2))/(2)`
`and" "k=(2at+2at')/(2)=a(t+t')=-(2a)/(t)` (Using (1))
`:." "t=-(2a)/(k)` (2)
Now, `h=(at^(2)+a(-t-(2)/(t))^(2))/(2)`
`rArr" "(2h)/(a)=(-(2a)/(k))^(2)+((2a)/(k)+(k)/(a))^(2)`
`rArr" "(2x)/(a)=(4a^(2))/(y^(2))+((2a)/(y)+(y)/(a))^(2)`
`rArr" "2axy^(2)=4a^(2)+(2a^(2)+y^(2))^(2)`, which is required locus.
Alternative method :
Equation of normal at point P(t) is
`y=-tx+2at+at^(3)` (1)
It meets the parabola at point Q.
Midpoint of PQ is R (h,k).
So, equation of chord PQ, whose midpoint is R(h,k), is
`ky-2a(x+h)=k^(2)-4ah" (Using "T=S_(1))` (2)
Equations (1) and (2) represent the same straight line.
`:." "(t)/(-2a)=(1)/(k)=(2at+at^(3))/(k^(2)-2ah)`
`rArr" "t=-(2a)/(k)and(1)/(k)=(2a(-(2a)/(k))+a(-(2a)/(k))^(3))/(k^(2)-2ah)`
`rArr" "(y^(2)-2ax)/(y)=(4a)/(y)-(8a^(4))/(y^(3))`
`rArr" "2axy^(2)=4a^(4)+(2a^(2)+y^(2))^(2)`, which is required locus.