Question

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

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

Answer 1

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

If line $y-\sqrt{3}x+3=0$ cuts parabola $y^2=x+2$ at $A$ and $B$

If $P(3,0)$ the $PA.PB$ is :

$y-\sqrt{3}x+3=0y=\sqrt{3}x-3\frac{y-0}{\frac{\sqrt{3}}{2}}=\frac{x-\sqrt{3}}{\frac{1}{2}}=rx=\frac{r}{2}+\sqrt{3}y=\frac{\sqrt{3}r}{2}$

Put in $y^2=x+2$

$\frac{3}{4}{r}^2=\frac{r}{2}+\sqrt{3}+2{3r}^2-2r-4(\sqrt{3}+2)=0PA.PB=\left|r_1{r}_2\right|=\left|\frac{c}{a}\right|=\frac{4(\sqrt{3}+2)}{3}$