At any point P on the parabola y2 – 2y - 4x + 5 = 0, a tangent is drawn which meets the directrix at Q. Find the locus of point R, which divides QP externally in the ratio 1/2 : 1.

Dear Bhanu,

 y2 – 2y - 4x + 5 = 0,
(y-1)^2 = 4(x-1)
Y^2 = 4X...............
4a = 4
a =1
The tangent of the parabola is given as:
y = mx + 1/m
thus at directric  its points will be: (-1, 1/m – m) as the eqn. of directrix is given by: x + a = 0
.i.e. x + 1 = .0

And as we know that the foot of the tangent on parabola is : (1/m^2,2/m).
Thus now simply we have a qstn. that 
a line segment having vertices located at  (1/m^2,2/m).  and (-1, 1/m -m) in the ratio 1/2:1
Let R be the point (h,k) from where it is externally divided in that ration.
Thus apply simple ratio formulaes and equate it to h and k respectively.
After that just eliminate m from h and k eqn. and get the realtion between h and k.
.
And lastly dont forget to put Y =y 1  and X = x- 1.
That is the answer.

Hope it helps!