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  • If <img alt="\mathrm{ P \equiv(\sqrt{3}, 0) and ~line ~y-\sqrt{3} x+3=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20P%20%5Cequiv%28%5Csqrt%7B3%7D%2C%200%29%20and%20%7

If \mathrm{ P \equiv(\sqrt{3}, 0) and ~line ~y-\sqrt{3} x+3=0} cuts the parabola \mathrm{ y^2=x+2}  at A and B, then \mathrm{ |P A . P B|} is equal to
 

Option: 1

\frac{4 \sqrt{3}}{2}


Option: 2

2\left(\frac{\sqrt{3}+2}{3}\right)


Option: 3

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


Option: 4

4\left(\frac{\sqrt{3}+2}{3}\right)


Answers (1)

best_answer

Given parabola is \mathrm{y^2=x+2 \quad \quad\dots(i)}
Given line is \mathrm{y=\sqrt{3} x-3 \quad \quad\dots(ii)}
\mathrm{P \equiv(\sqrt{3}, 0) }
A B makes an angle of \mathrm{60^{\circ}}  with the positive direction of x-axis. Coordinates of any point on this line may be taken as

\mathrm{\left(\sqrt{3}+r \cos 60^{\circ}, 0+r \sin 60^{\circ}\right) \text { i.e. }\left(\sqrt{3}+\frac{r}{2}, \frac{r \sqrt{3}}{2}\right) \text {. } }
If this point lies on (i), then

\mathrm{\frac{3}{4} r^2=\sqrt{3}+\frac{r}{2}+2 \quad \text { or } \quad 3 r^2=4 \sqrt{3}+2 r+8}
or
\mathrm{3 r^2-2 r-4(2+\sqrt{3})=0} \quad \quad \dots(iii)
Let \mathrm{r_1}  and \mathrm{r_2}  be the roots of equation (iii), then

\mathrm{r_1 r_2=-\frac{4(2+\sqrt{3})}{3} }
Now \mathrm{|P A \cdot P B|=\left|r_1 r_2\right|=\frac{4}{3}(2+\sqrt{3})}

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Gaurav

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