IIT JEE Advanced 2006 Paper 1

It is given that
y2 = 4x
The equation of normal is
y = mx – 2m – m3 (1)
which is passing through the point (3, 0). Therefore, Eq. (1) becomes
0 = 3m – 2m – m3
m3 – m = 0 ⇒ m = 0, −1, 1
That is, ; hence, we have the points P(0, 0), Q(1, 2) and R(1, −2) as shown in the following figure:

(A) The area of triangle PQR is

1

2

× 1 × 4 = 2
Therefore, (A)→(P).
(B) The radius of circumcircle is
R =

abc

4×Area‌of‌Δ‌P‌Q‌R

=

4√5√5

4×2

=

5

2

Therefore, (B)→(Q).
(C) The centroid of ΔPQR is
(

0+1+1

3

,

0+2−2

3

) = (

2

3

,0)
Therefore, (C)→(S).
(D) The circumradius of ΔPQR is 5/2 and the circumcentre of
ΔPQR is (

5

2

,0)

Therefore, (D)→(R).