If a tangent to the parabola `y^2=4a x` meets the x-axis at `T` and intersects the tangents at vertex `A` at `P ,` and rectangle `T A P Q` is completed, then find the locus of point `Qdot`

Explanation

Let `P(at^(2), 2at)` be any point on the parabola `y^(2) = 4ax`. Then tangent at `P(at^(2), 2at)` is ty = x + `at^(2)`
Since tangent meet the axis of parabola in T and tangent at the vertex in Y.

`therefore` Co-ordinates of T and Y are `(-at^(2),0)` and (0,at) respectively.
Let co-ordinates of G be `(x_(1), y_(1))`.
Since TAYG is rectangle.
`therefore` Mid-points of diagonals TY and GA is same
`!(x_(1)+0)/2=(-at^(2)+0)/2`
`!x_(1)=-at^(2)` ...(i)
and `(y_(1)+0)/2=(0+at)/2`
Eliminating t from (i) and (ii) then we get `x_(1)=-a((y_(1))/a)` or
`y_(1)^(2)=-ax_(1)`
or `y_(1)^(2)+ax_(1)=0`
`therefore` The locus of G`(x_(1),y_(1))` is `y^(2)+ax=0`