Question

A line is drawn from A (–2, 0) to intersect the curve y² = 4x in P and Q in the first quadrant such that$\frac{1}{AP} + \frac{1}{AQ} < \frac{1}{4}$, then slope of the line is always-

• A. <$\sqrt{3}$
• B. >$\sqrt{3}$
• C. ³$\sqrt{3}$
• D. None of these

Answer 1

Answer: (B)

>$\sqrt{3}$

Answer 2

Let P (–2 + r cos q, r sin q) and P lies on parabola

Ž r² sin q – 4(–2 + r cos q) = 0

Ž r₁ + r₂ = $\frac{4\cos\theta}{\sin^{2}\theta}$ø r₁r₂ = $\frac{8}{\sin^{2}\theta}$

\ $\frac{r_{1} + r_{2}}{r_{1}r_{2}}$ = $\frac{1}{AP} + \frac{1}{AQ}$ Ž cos q ¹ $\frac{1}{2}$

Ž tan q > $\sqrt{3}$. (because cos q is decreasing and tan q is increasing in (0, p/2)).

Ž m >$\sqrt{3}$

Hence (2) is correct answer.