  Question

# If the tangents at P and Q in the parabola meet in T, then which of the following statements are correct. 1. TP and TQ subtends equal angle at the focus S 2. ST2 = SP.SQ Only 1 Both 1 2 None of these Only 2

Solution

## The correct option is B Both 1 2 Let's take a standard parabola y2=4ax and draw tangents at P(at21,2at1) and Q(at22,2at2)... Which meet at T. We know the point of intersection of tangent is T(at1t2,a(t1+t2)) We can also calculate coordinates of T by calculating tangents at P & Q and their point of intersection is point T. Tangent at P y(2at1)=2a(x+at21)                   yt1=x+at21                              ............(1) Similarly tangent at Q yt2=x+at22                              ............(2) Solving equation (1) & (2) We get, x=at1t2              y=a(t1+t2) Coordinates of T(at1t2,a(t1+t2) Statement 1 To,check α=β, or check that T lies on the angle bisector of the ∠PSQ. i.e., perpendicular distance of T from the line SP is equal to the perpendicular of T from SQ. Equation of SP y=2at2at21−a(x−a) 2t1x−(t21−1)y−2at1=0 P1=2at21t2−(t21−1)a(t1+t2)−2at1√(t21−1)2+4t21 P1=a|t1−t2| Similarly equation of SQ y=2at2at22−a(x−a) 2t2x−(t22−1)−2at2=0 Perpendicular distance of SQ from T P2=2at22t1−(t22−1)a(t1+t2)−2at2√(t22−1)2+4t22 So, we can say that α=β Statement 1 is correct. Statement 2: ST2=(a−at1t2)2+(a(t1+t2)−0)2           =a2(1−t1.t2)2+a2(t1+t2)2           =a2[1+t21t22+t21+t22] SP=√(at21−a)2+(2at1−0)2=√a2(t21−1)+4a2t21      =a√t41+1−2t21+4t21=a√(t21+1)2=a(t21+1) SQ=√(at22−a)2+(2at1−0)2=√a2(t22+1)+4a2t22      =a√t42+1−2t22+4t22=a√(t22+1)2=a(t22+1) SP.SQ=a(t21+1)×a(t22+1)                 =a2[t21t22+t21+t22+1] SP.SQ=a2[1+t21t22+t21+t22]=ST2 Statement 2 is correct Both the statement are correct.  Suggest corrections   