Question

If the line $y=x\sqrt{3}-3$ cuts the parabola $y^2=x+2$ at $P$ and $Q$ and if $A$ be the points $(\sqrt{3},0)$ , then $AP.AQ$ is

• A. $\frac{2}{3}(\sqrt{3}+2)$
• B. $\frac{4}{3}(\sqrt{3}+2)$
• C. $\frac{4}{3}(2-\sqrt{3})$
• D. $\frac{4}{6-3\sqrt{3}}$

Answer 1

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

Parabola: $y^2=x+2..........\left(1\right)$

Line: $y=x\sqrt{3}-3...........\left(2\right)$

$A(\sqrt{3},0)$ lies on $\left(2\right)$ and at $x-axis$,

$P$ with respect to $A,$ is $P=(\sqrt{3}+rcos60°,rsin60°)$

$=(\sqrt{3}+\frac{r}{2},\frac{\sqrt{3r}}{2})$ (where $r=PA)$

$P$ is on Parabola,

$=>(\frac{\sqrt{3}r}{2}{)}^2=\sqrt{3}+\frac{r}{2}+2$

$=>\frac{{3r}^2}{4}=\sqrt{3}+2+\frac{r}{2}$

$=>3r^2-2r+4(-\sqrt{3}-2)=0$

Is quadratic equation, having two roots say $r_1$ and $r_2$ which are lengths of $PA$ and $AQ$

$AP.AQ=r_1.r_2$

$AP.AQ=\frac{4}{3}(\sqrt{3}+2).$