If the line `y-sqrt(3)x+3=0` cut the parabola `y^2=x+2` at `P` and `Q` , then `A PdotA Q` is equal to [where `A=(sqrt(3),o)]` `(2(sqrt(3)+2))/3` (b) `(4sqrt(3))/2` `(4(2-sqrt(2)))/3` (d) `(4(sqrt(3)+2))/3`

Options

A. `(2(sqrt(3)+2))/(3)`
B. `(4sqrt(3))/(2)`
C. `(4(2-sqrt(2))/(3))`
D. `(4(sqrt(3+2))/(3))`

Answer

D

Explanation

(4) `y-sqrt(3)x+3=0` can be rewritten as
`(y-0)/(sqrt(3)//2)=(x-sqrt(2))/(1//2)=r` (1)
On solving (1) with the parabola `y^(2)=x+2`, we get
`(3r^(2))/(4)=(r)/(2)+sqrt(3)+2`
`or3r^(2)-2r-(4sqrt(3)+8)=0`
Hence, `AP*AQ=|r_(1)r_(2)|=(4(sqrt(3)+8))/(3)` (Product of roots)