If the line`y-sqrt(3)x + 3=0` cuts the parabola `y^2=x + 2` at A and B, then find the value of PA.PB(where `P=(sqrt(3),0)`

If the line y-sqrt3x+3=0 cuts the parabola y^2=x+2 at P and Q then AP.AQ is equal to

If the line y-sqrt(3)x+3=0 cuts the parabola y^(2)=x+2 at P and Q then AP.AQ is equal to

If the line y-sqrt3x+3=0 cuts the parabola y^2=x+2 at P and Q then AP.AQ is equal to

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

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),0)] (a) (2(sqrt(3)+2))/3 (b) (4sqrt(3))/2 (c) (4(2-sqrt(2)))/3 (d) (4(sqrt(3)+2))/3

If the line y-sqrt(3)x+3=0 cuts the curve y^(2)=x+2 at A and B and point on the line P is (sqrt(3), 0) then PA. PB =

If the line y=sqrt(3)x+3=0 cuts the curve y^(2)=x+2 at A ,B and the point P(sqrt(3) ,0) .Then |PA*PB| =

The line 2y=3x+12 cuts the parabola 4y=3x^(2) . Where does the line cut the parabola ?