InterviewSolution
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InterviewSolution
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Find the length of the normal chord which subtends an angle of `90^@` at the vertex of the parabola `y^2=4x` . |
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Answer» Let `PQ` is the common chord. Then, at any point coorinates of `P` and `Q` are `(t1^2,2t1)` and `(t2^2,2t2)` As the chord subtends an angle of `90^@`, relation between t1 and t2 will be, `t2=-t1-2/(t1)` `-> Eq(1)` If O is the origin, then OP and OQ are perpendicular to each other. In that case, both their slopes multiplication will be -1. Thus, `(2t1)/(t1^2)**(2t2)/(t2^2)=-1` `=>t1.t2=-4 ->Eq(2)` Putting values of t2 from Eq(1) `=>t1(-t1-2/(t1))=-4` `=>-t1^3-2t1=-4t1` `=>t1^3-2t1=0` Solving, above, we get, `t1=0` or `t1=sqrt(2)` As, t1=0 is not possiblee, so `t1 = sqrt(2)` Putting value of t1 in Eq(2), we get, `t2 = -2sqrt(2)` So, coordinates of chord are`P(2,2sqrt(2))` and `Q(8,-4sqrt(2))` So, length of PQ will be `sqrt((6sqrt(2))^2+6^2))= sqrt(108) = 6sqrt(3)` |
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