IIT JEE Advanced 2007 Paper 1

See the figure shown here. We have the circle $x^2+y^2$ = 9 and parabola $y^2$ = 8x Points P and Q are the point of intersection of the circle and the parabola. Therefore, we get P (1 , $2\sqrt{2}$) Q (1 , - $2\sqrt{2}$) 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 = $\frac{1}{2} \times 4\sqrt{2} \times 8$ = $16\sqrt{2}$ Area of PQS = $\frac{1}{2} \times 4\sqrt{2} \times 2$ = $4\sqrt{2}$ The ratio of the area of ΔPQS and ΔPQR is 1:4.