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Q.
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 $P$ is $\sqrt{3} , 0$)
Conic Sections
Solution:
$y-\sqrt{3}x+3 = 0$ can be written as
$\frac{ y-0}{\frac{\sqrt{3}}{2}} =\frac{ x-\sqrt{3}}{\frac{1}{2}} = r \quad...\left(1\right)$
Solving $\left(1\right)$ with the parabola $y^{2}=x+2 $
we get $\frac{3r^{2}}{4} = \frac{r}{2}+\sqrt{3} +2$
$\Rightarrow 3r^{2}-2r-\left(4\sqrt{3} +8\right) = 0$
$ \therefore PA.PB = \left|r_{1}, r_{2}\right| = \frac{4\sqrt{3}+8}{3} $
$= \frac{4\left(\sqrt{3} +2\right)}{3}$