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Let S be the focus of the parabola `y^2=8x` and let PQ be the common chord of the circle `x^2+y^2-2x-4y=0` and the given parabola. The area of the triangle PQS is -A. 4B. 3C. 2D. 8 |
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Answer» Correct Answer - A The parametric equations of the parabola are `x=2t^(2), y=4t`, Putting `x=2t^(2)` and y=4t in `x^(2)=y^(2)-2x-4y=0,` we get `4t^(4)+156t^(2)-4t^(2)-16t=0` `rArr" "4t^(2)+12t^(2)-16t=0` `rArr" "4t(t^(3)+3t-4)=0` `rArr" "t(t-1)(t^(2)+t+4)=0rArrt=0, t=1" "[because t^(2)+t+4ne0]` Thus, the coordinates of points of intersection of the circle and the parabola are O(0, 0) and P(2, 4). Clearly, these are diametrically opposite points on the circle. The coordinates of the focus S of the parabola are (2, 0) which lies on the circle. So, OSP is a right trangle. `:." Area of "DeltaOSP=1/2xxOSxxSP=1/2xx2xx4=4` |
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