IIT JEE Advanced 2007 Paper 1

See the figure shown here.

We have the circle
x2+y2 = 9
and parabola
y2 = 8x
Points P and Q are the point of intersection of the circle and the parabola. Therefore, we get
P (1 , 2√2)
Q (1 , - 2√2)
The equation of PQ is x =1. Now, let R = (α, 0) and PQ is the chord of contact of point R about the circle. Therefore, the line
α(x) + 0(β) = 9
is similar to PQ. This implies that
α = 9
R (9, 0)
Now, let s(β, 0) and PQ is chord of contact of s about the parabola.
Therefore, the line
y(0) = 4(x + β)
is similar to PQ. This implies that
β = - 1
s (- 1 , 0)
Now,
Area of PQR =

1

2

×4√2×8 = 16√2
Area of PQS =

1

2

×4√2×2 = 4√2
The ratio of the area of ΔPQS and ΔPQR is 1:4.