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
<13
>13
>2
>3
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=0
Let P and Q be distances r1 and r2 from A, then
r1+r2=4cosθsin2θ, r1r2=8sin2θ
1AP+1AQ=1r1+1r2=r1+r2r1r2=cosθ2
Given 1AP+1AQ<14
cosθ2<14⇒cosθ<12⇒tanθ>3