A tangent and a normal are drawn at the point P(2,−4) on the parabola y2=8x, which meet the directrix of the parabola at the points A and B respectively. If Q(a,b) is a point such that AQBP is a square, then 2a+b is equal to :
(A)-12
(B)-18
(C)-16✓ Correct
(D)-20
Answer: (C)
Solution
The parabola y2=8x has 4a=8, so its directrix is x=−2. The tangent at P(2,−4) is y(−4)=4(x+2), i.e. x+y+2=0, meeting x=−2 at A(−2,0). The normal is perpendicular to it, with slope 1: y+4=x−2, i.e. x−y−6=0, meeting x=−2 at B(−2,−8). In the square AQBP the diagonals AB and PQ bisect each other, so the midpoint of PQ is that of AB, namely (−2,−4). With P(2,−4) this gives Q(−6,−4), hence 2a+b=2(−6)+(−4)=−16.
JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option
