Question

If the line $y-\sqrt{3}x+3=0$ cuts the parabola $y^2=x+2$ at P and Q, then AP $\cdot$ AQ is equal to [where A $\equiv$ ($\sqrt{3}$, 0)]

• A. $\frac{2(\sqrt{3}+2)}{3}$
• B. $\frac{4\sqrt{3}}{2}$
• C. $\frac{4(2-\sqrt{2})}{3}$
• D. $\frac{4(\sqrt{3}+2)}{3}$

Answer 1

The correct option is D $\frac{4(\sqrt{3}+2)}{3}$

$y-\sqrt{3}x+3=0$ can be rewritten as
$\frac{y-0}{\sqrt{3}/2}=\frac{x-\sqrt{3}}{1/2}=r$ ......... (i)

A is a point lying on the line $y-\sqrt{3}x+3=0$
On solving equation (i) with the parabola $y^2=x+2$
$\frac{3r^2}{4}=\frac{r}{2}+\sqrt{3}+2$
$\Rightarrow 3r^2-2r-(4\sqrt{3}+8)=0$
$\Rightarrow AP\cdot AQ=|r_1{r}_2|$
$=\frac{4(\sqrt{3}+2)}{3}$ (product of roots)