Tangent is drawn to parabola $y^{2} - 2 y - 4 x + 5 = 0$ at a point P, which cuts the directrix at the point Q. A point R is such that it divides QP externally in the ratio $\frac{1}{2}$ :1 .The locus of the point R is:

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${(x + 1)}^{2} (y - 1) - 4 = 0$

$(x - 1) {(y + 1)}^{2} + 4 = 0$

$2 {(x + 1)}^{2} (3 y - 1) + 3 = 0$

$(x + 1) {(y - 1)}^{2} + 4 = 0$

answer is D.

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Detailed Solution

Parabola ${(y - 1)}^{2} = 4 (x - 1) \to$ Directrix: y-axis x=0
Any point on parabola: $p (t^{2} + 1, 2 t + 1)$
Eq.of tangent at P: $y (2 t + 1) - (y + 2 t + 1) - 2 (x + t^{2} + 1) + 5 = 0$
$\Rightarrow x - t y + (t^{2} + t - 1) = 0$
Tangent meets directrix at Q $(0, \frac{t^{2} + t - 1}{t})$
Let R = (h, k); Q is mid-point of PR
$\frac{h + t^{2} + 1}{2} = 0 a n d \frac{k + 2 t + 1}{2} = \frac{t^{2} + t - 1}{t}$
$\Rightarrow t^{2} = - 1 - h & 2 = t (1 - k)$
Eliminating‘t’, $4 = (- 1 - h) {(1 - k)}^{2}$
Locus of R $(x + 1) {(y - 1)}^{2} + 4 = 0$