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Answer

`(x+1)(y-1)^(2)+4=0`

Explanation

We have parabola `(y-1)^(2)=4(x-1)`.
Here, directrix is y-axis, i.e.,x=0.
Let point P on the parabola be `(t^(2)+1,2t+1)`.
So, eqution of tangent to parabola at point P is
`t(y-1)=(x-1)+t^(2)` (1)
`or" "x-ty+(t^(2)+t-1)=0`
It meets directrix at `Q(0,(t^(2)+t-1)/(t))`.
Now, Q is midpoint of RP.
`:." "(h+t^(2)+1)/(2)=0and(k+2t+1)/(2)=(t^(2)+t-1)/(t)`
`rArr" "t^(2)=-1-handkt+2t^(2)+t=2t-2`
`rArr" "t^(2)=-1-hand2=t(1-k)`
Eliminating t, we get
`4=(-1-h)(1-k)^(2)`
`or" "(x+1)(y-1)^(2)+=0`,
which is the locus point R.