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Answer

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

Explanation

Given equation can be rewritten as
`(y - 1)^(2) = 4 (x - 1)`, whose parametric coordinates are
`x - 1 = t^(2)` and `y - 1 = 2t`
i.e., `P (1 + t^(2), 1 + 2t)`
`:.` Equation of tangent as P is
`t (y - 1) = x - 1 + t^(2)` Which meets the directrix x = 0 at Q.
`implies y = 1 + t - (1)/(t)` or `Q (0,1 + t - (1)/(t))`
Let R (h,k) which divides QP externally in the ratio `(1)/(2) : 1` or Q is mid point of RP
`implies 0 = (h + t^(2) + 1)/(2)` or `t^(2) = - (h + 1)` .....(i)
and `1 + t - (1)/(t) = (k + 2t + 1)/(2)` or `t = (2)/(1 - k)`....(ii) ,
From Eqs. (i) and (ii), `(4)/((1 - k)^(2)) + (h + 1) = 0`
or `(k - 1)^(2) (h + 1) + 4 = 0`