Question

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

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

Answer 1

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

Line $=y-\sqrt{3}x+3=0$

$\begin{array}{c}y=\sqrt{3}x-3\\ m=\sqrt{3}=\tan \theta \\ \theta =\frac{\pi }{3}={60}^{\circ }\to \left(i\right)\end{array}$

Parametric form of line

$\begin{array}{c}=\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }=r\\ \therefore x=x_1+r\cos \theta \\ y=y_1+r\sin \theta \end{array}$

As we know $P\left(\sqrt{3},0\right)$

$\therefore \begin{array}{c}x=\sqrt{3}+r\cos \theta \\ y=0+r\sin \theta \end{array}\}lieonparabola{y}^2=x+2$

$\begin{array}{c}\therefore {\left(r\sin \theta \right)}^2=\sqrt{3}+r\cos \theta +2\\ r^2{\sin }^2\theta =\sqrt{3}+r\cos \theta +2\\ r^2{\sin }^2\theta -r\cos \theta -\sqrt{3}-2=0\begin{array}{c}{r}_1=PA\\ r_2=PB\end{array}\\ r_1{r}_2=\frac{-\sqrt{3}-2}{{\sin }^2\theta }\to \left(ii\right)\\ from\left(i\right)\theta ={60}^{\circ }\sin {60}^{\circ }=\frac{\sqrt{3}}{2}{\sin }^{2}60=\frac{3}{4}\\ from\left(ii\right)\frac{-\sqrt{3}-2}{\frac{3}{4}}=\frac{-4\left(\sqrt{3}+2\right)}{3}\\ \therefore \left|PA.PB\right|=\left|r_1{r}_2\right|=\frac{4\left(\sqrt{3}+2\right)}{3}\end{array}$

Hence ans is A.