y2 = 4x Point P lies on normal to parabola passing through centre of circle ∴ y + tx = 2t + t3 ... (1) (Equation of normal at P to parabola) ⇒ 8 + 2t = 2t + t3 ∴ t = 2 ⇒ P (4 , 4) SP = √(4−2)2+(4−8)2 ∴ SP = 2√5 And SQ = 2 ⇒ PQ = 2√5 - 2 ∴
SQ
QP
=
1
√5−1
=
√5+1
4
To find x intercept put y = 0 in (1) ⇒ x = 2 + t2 ⇒ x = 6 (As t = 2) ∵ Slope of common normal = - t = - 2 Slope of tangent at Q on circle =