Question

If the line $y-\sqrt{3}x+3=0$ cuts the parabola $y^2=x+2$ at $A$ and $B$, then $PA.PB$is equal to [ where, $P=\left(\sqrt{3},0\right)$ ]

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

Answer 1

The correct option is B

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

Explanation of the correct option:

Let's plot the curves.

Given: $P=\left(\sqrt{3},0\right)$

The equation of line $AB$ is $\frac{\left[x-\sqrt{3}\right]}{\cos \left(60°\right)}=\frac{\left[y-0\right]}{\sin \left(60°\right)}=r$

$$\begin{gathered} \Rightarrow x=\sqrt{3}+\frac{r}{2} \\ \Rightarrow y=\frac{r\sqrt{3}}{2} \end{gathered}$$

Therefore, point $\left(\sqrt{3}+\frac{r}{2},\frac{r\sqrt{3}}{2}\right)$ lies on $y^2=x+2$

$$\begin{gathered} \Rightarrow \frac{3r^2}{4}=\sqrt{3}+\frac{r}{2}+2 \\ \Rightarrow \frac{3r^2}{4}-\frac{r}{2}-\left(2+\sqrt{3}\right)=0 \end{gathered}$$

Let the roots be $r_1$ and $r_2$, then the product

$\begin{array}{rcl}{r}_1\times r_2& =& PA.PB\\ & =& \left|\frac{-\left(2+\sqrt{3}\right)}{{\displaystyle \frac{3}{4}}}\right|\\ & =& \frac{4\left(2+\sqrt{3}\right)}{3}\end{array}$

Hence option $\left(B\right)$ is the correct option.