JEE Advanced2012MathematicsParabolaActual
Let S be the focus of the parabola y²=8 x and let P Q be the common chord of the circle x²+y²-2 x-4 y=0 and the given parabola. The area of the triangle P Q S is
Correct answer
4
Step-by-step solution
Given parabola y²=8 x and circle x²+y²-2 x-4 y=0 pass through the origin One end of common chord PQ is origin. Say P (0,0) Let Q be the point (2 t², 4 t ) , then it will satisfy the equation of circle. 4 t⁴+16 t²-4 t²-16 t=0 t⁴+3 t²-4 t=0 t (t³+3 t-4 )=0 t(t-1) (t²+t-4 )=0 t=0 or 1 For t=0 , we get point P , therefore t=1 gives point Q as (2,4) . We also observe here that P (0,0) and Q (2,4) are end points of diameter of the given circle and focus of the parabola is the point S(2,0) . area ( PQS )= 1 2 P S Q S= 1 2