Let a line y=mx(m>0) intersect the parabola y2=x at a point P, other than the origin. The tangent to it at P meet the x− axis at point Q. If area △OPQ=4 sq.units, then 2m is equal to
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Solution
Given parabola is y2=x Let the co-ordinates of P be (t24,t2) Putting the P in y=mx, we get m=2t Equation of tangent at P is yt=x+t24 Now coordinates of Q≡(−t24,0)
Area of △OPQ=4 ⇒4=12×base×height⇒12×OQ×PM=4⇒12×t24×t2=4⇒t=4⇒m=12∴2m=1