A line is drawn from A(-2,0) to intersect the curve y2=4x at P and Q in the first  quadrant such that 1AP+1AQ<14 . Then the slope of the line is always

Let any point at a distance r from A on the parabola is (-2+rcosθ, rsinθ), P lies on y2=4x ,⇒r2sin2θ  =4(-2+rcos  θ  ), r2sin2θ  -4rcosθ+8=0Let P and Q be distances r1 and r2 from A, then r1+r2=4cosθsin2θ,  r1r2=8sin2θ1AP+1AQ=1r1+1r2=r1+r2r1r2=cosθ2 Given 1AP+1AQ<14cosθ2<14⇒cosθ<12⇒tanθ>3