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Question: If the line y –\(\sqrt{3}\)x + 3 = 0 cuts the parabola y<sup>2</sup> = x + 2 at A and B, and if P ≡ ...

If the line y –3\sqrt{3}x + 3 = 0 cuts the parabola y2 = x + 2 at A and B, and if P ≡ (3\sqrt{3}, 0), then PA.PB is equal to

A

2(3+2)3\frac{2\left( \sqrt{3} + 2 \right)}{3}

B

432\frac{4\sqrt{3}}{2}

C

4(2−3)3\frac{4\left( 2 - \sqrt{3} \right)}{3}

D

4(3+2)3\frac{4\left( \sqrt{3} + 2 \right)}{3}

Answer

4(3+2)3\frac{4\left( \sqrt{3} + 2 \right)}{3}

Explanation

Solution

We may write y – 3\sqrt{3} x + 3 = 0 as y−03/2=x−31/2=r\frac{y - 0}{\sqrt{3}/2} = \frac{x - \sqrt{3}}{1/2} = r. ...(i)

Substituting x = 3+r2,y=32r\sqrt{3} + \frac{r}{2},y = \frac{\sqrt{3}}{2}r in y2 = x + 2, we get

3r24=r2+3+2\frac{3r^{2}}{4} = \frac{r}{2} + \sqrt{3} + 2⇒ 3r2- 2r – (43\sqrt{3} + 8) = 0

⇒ PA . PB = |r1. r2| = (43+8)3\frac{\left( 4\sqrt{3} + 8 \right)}{3} =4(3+2)3\frac{4\left( \sqrt{3} + 2 \right)}{3}