Question

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

Answer 1

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

The point $A(\sqrt{3},0)$ lies on the given line whose slop is $\sqrt{3}$
$\therefore tan\theta =\sqrt{3}=tan{60}^{\circ }\therefore \theta ={60}^{\circ }$
Equation on line through A is
$\frac{x-\sqrt{3}}{cos{60}^{\circ }}=\frac{y-0}{sin{60}^{\circ }}=r$
Where r is the distance of any point from A. it intersects the parabola $y^2=x+2$
$\therefore {\left(\frac{r\sqrt{3}}{2}\right)}^2=\frac{r}{2}+\sqrt{3}+2\Rightarrow 3r^2+2r+4(2+\sqrt{3})=0$
$\therefore r_1{r}_2=\frac{4(2+\sqrt{3})}{3}=AP.AQ\Rightarrow \left(b\right)$