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  • If the line  cuts the parabola <img alt="\mathrm{y^2=x+2}

If the line \mathrm{y=\sqrt{3} x-3 } cuts the parabola \mathrm{y^2=x+2} at \mathrm{A} and \mathrm{B}, then find \mathrm{P A \cdot P B}, where \mathrm{P} is \mathrm{(\sqrt{3}, 0).}

Option: 1

\mathrm{\frac{4(2+\sqrt{3})}{3}}


Option: 2

\mathrm{\frac{4(2-\sqrt{3})}{3}}


Option: 3

\mathrm{\frac{(2+\sqrt{3})}{2}}


Option: 4

\mathrm{\frac{(2-\sqrt{3})}{2}}


Answers (1)

best_answer

Equation of \mathrm{A B} in parametric form
\mathrm{ \frac{x-\sqrt{3}}{\frac{1}{2}}=\frac{y-0}{\frac{\sqrt{3}}{2}}=r }
\mathrm{\Rightarrow} Any point on line is \mathrm{\left(\sqrt{3}+\frac{r}{2}, \frac{\sqrt{3}}{2} r\right)} and satisfies the equation of parabola \mathrm{\mathrm{y}^2=\mathrm{x}+2.}

\mathrm{ \begin{aligned} & \Rightarrow \frac{3}{4} r^2=\sqrt{3}+\frac{r}{2}+2 \Rightarrow 3 r^2=4 \sqrt{3}+2 r+8 \\ & \Rightarrow 3 r^2-2 r-4(2+\sqrt{3})=0 \\ & \text { PA. PB }=\left|r_1 \cdot r_2\right|=\left(\frac{4(2+\sqrt{3})}{3}\right) \end{aligned} }

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Rakesh

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