$A line is drawn from A (- 2, 0) to intersect the curve y^{2} = 4 x at P and Q in the first$ $quadrant such that \frac{1}{A P} + \frac{1}{A Q} < \frac{1}{4} . Then the slope of the line is always$

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$< \frac{1}{\sqrt{3}}$

$> \frac{1}{\sqrt{3}}$

$> \sqrt{2}$

$> \sqrt{3}$

answer is D.

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Detailed Solution

Let any point at a distance r from A on the parabola is (-2+rcos $θ$ , rsin $θ$ ), P lies on $y^{2} = 4 x$ ,

$\Rightarrow r^{2} \sin^{2} θ = 4 (- 2 + r \cos θ),$

$r^{2} \sin^{2} θ - 4 r \cos θ + 8 = 0$

Let P and Q be distances $r_{1}$ and $r_{2}$ from A, then

$r_{1} + r_{2} = \frac{4 \cos θ}{\sin^{2} θ}, r_{1} r_{2} = \frac{8}{\sin^{2} θ}$

$\frac{1}{A P} + \frac{1}{A Q} = \frac{1}{r_{1}} + \frac{1}{r_{2}} = \frac{r_{1} + r_{2}}{r_{1} r_{2}} = \frac{\cos θ}{2}$

$Given \frac{1}{A P} + \frac{1}{A Q} < \frac{1}{4}$

$\frac{\cos θ}{2} < \frac{1}{4} \Rightarrow \cos θ < \frac{1}{2} \Rightarrow \tan θ > \sqrt{3}$

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