JEE Advanced Model Paper 1 with solutions for online practice

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 =

1

2