Question

If the line $y=\sqrt{3}x-3$ cuts the parabola $y^2=x+2$ at $P$ and $Q$ and 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

Answer: (B)

$-\frac{4}{3}(\sqrt{3}+2)$

${\left(\sqrt{3}(x-\sqrt{3})\right)}^2=x+2$
$3(x^2-2\sqrt{3x}+3)=x+2$
$3x^2-6\sqrt{3x}-x+7=0$
$PA.AQ$$=\sqrt{{(x_1-p\sqrt{3})}^2+y_1^2}\sqrt{{(x_2-\sqrt{3})}^2+y_2^2}$
$=\sqrt{{(x_1-\sqrt{3})}^2+(\sqrt{3}(x_2-\sqrt{3}){)}^2}\sqrt{(x_2-\sqrt{3})+{\left(\sqrt{3}(x_2-\sqrt{3})\right)}^2}$
$=4(x_1-\sqrt{3})(x_2-\sqrt{3})$
$=4(x_1{x}_2-\sqrt{3}(x_1+x_2)+3)$
$=4(\frac{7}{3}-\frac{\sqrt{3}(6\sqrt{3}+1)}{3}+3)$
$=\frac{-4}{3}(\sqrt{3}+2)$