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Question

# If a point P on the axis of the parabola y2=4x is taken such that the point is at shortest distance from the circle x2+y2+2x−2√2y+2=0.Common tangents are drawn to the circle and the parabola from P.If the area of the ΔPAB is √a sq. units where A and B are the points of contact on two distinct tangents from P on circle and parabola respectively, then a=

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Solution

## Axis of the parabola y2=4x is y=0, and the point which is at shortest distance from the circle. x2+y2+2x−2√2y+2=0 is (−1,0) Since, (−1,0) lies on the directrix of the parabola hence tangents to the parabola are y=x+1 and y=−x−1 and these lines are also the tangents to the given circle. Here, △PAB is possible two ways one shown by using PAB and other by using PA1B1. Length of PB=PB1=1, also A(1,2) and A1(1,−2) as A and A1 are the endpoints of the latusrectum of the parabola because latus rectum is only focal chord that subtends rightangle at axis of the parabola where it meets with directrix. Here, △PAB and △PA1B1 are right angle triangles. So, area of △PAB=12×PA×PB=12×1×2√2=√2sq.unit Also, area of (△PA1B1)=12×PA1×PB1=12×1×2√2=√2 sq unit Hence, area of △PAB=√2 sq unit

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