`P(t_1)` and `Q(t_2)` are the point `t_1a n dt_2` on the parabola `y^2=4a x` . The normals at `Pa n dQ` meet on the parabola. Show that the middle point `P Q` lies on the parabola `y^2=2a(x+2a)dot`

Explanation

The equation of normal to `y^(2)=4ax` is
`y=-tx+2at+at^(3)`
It passes through the (h,k). Then
`k=-tx+2at+at^(3)`
`or" "at^(3)+(2a-h)t-k=0`
Since normal at `P(at_(1)^(2),2at_(1))andQ(at_(2)^(2),2at_(2))` passes through the point `R(at_(3)^(2),2at_(3))` on the parabola,
`t_(1)t_(2)t_(3)=(k)/(a)=(2at_(3))/(a)`
`:." "t_(1)t_(2)=2`
Also , if `(x_(1),y_(1))` is the midpoint of PQ, then
`x_(1)=(1)/(2)(at_(1)^(2)+at_(2)^(2))andy_(1)=(1)/(2)(2at_(1)+2at_(2))` (2)
`:." "(t_(1)+t_(2))^(2)=((y_(1))/(a))^(2)`
`or" "((y_(1))/(a))^(2)=t_(1)^(2)+t_(2)^(2)+2t_(1)t_(2)=(2x_(1))/(a)+4`
`or" "y_(1)^(2)=2a(x_(1)+2a)`
Hence, the locus of `(x_(1),y_(1))` is `y^(2)=2a(x+2a)`.