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-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=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 sqrt(2y)-2x+8a=0 is a normal to the parabola y^(2)=4ax at P ,then P=

The line sqrt(2)y-2x+8a=0 is a normal to the parabola y^(2)=4ax at P .then P =

If the line x-3y+k=0 touches the parabola 3y^(2)=4x then the value of k is

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

If the line " 4y-3x-8=0 " cuts the parabola " x^(2)+y-4=0 " at "A" and "B" .and " |PA.PB| " is equal to " (k)/(8) " where " P=(0,2) " then the value of "k" is.

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