Question

If the line $y-\sqrt{3}x+3=0$ cuts the curve $y^2=x+2$ at $A$ and $B,P$ is a point on the line whose ordinate is $0.$ Then $|PA.PB|=$

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

Answer 1

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

$y=\sqrt{3}x-3$
So, Point $P\equiv (\sqrt{3},0)$

Let $PA=r$, inclination of $PA=\pi /3$
Using parametric form coordinates of
$A\equiv (\sqrt{3}+r\cos \frac{\pi }{3},0+r\sin \frac{\pi }{3})=(\sqrt{3}+\frac{r}{2},\frac{r\sqrt{3}}{2})$
$\because$A lies on curve
$y^2=x+2$
$\Rightarrow {\left(\frac{r\sqrt{3}}{2}\right)}^2=\sqrt{3}+\frac{r}{2}+2$
$\Rightarrow 3r^2=2r+4(2+\sqrt{3})$
$\Rightarrow 3r^2-2r-4(2+\sqrt{3})=0$
Which is a quadratic in $r$. Roots $r_1$and $r_2$ indicates $PA\text{and}PB$
So, $PA.PB=\frac{4(2+\sqrt{3})}{3}$