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JEE Advanced 2016 Paper 1 · Mathematics · Conic Sections · hard

The normal to the parabola y^2=4x at the point P(4,4) meets the parabola again at Q. The area of the region bounded by the parabola and the chord PQ…

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The normal to the parabola y2=4xy^2=4x at the point P(4,4)P(4,4) meets the parabola again at QQ. The area of the region bounded by the parabola and the chord PQPQ is
  1. A. 643\dfrac{64}{3}
  2. B. 1256\dfrac{125}{6}
  3. C. 1003\dfrac{100}{3}
  4. D. 1253\dfrac{125}{3}
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Answer: (D) 1253\dfrac{125}{3}

Differentiating y2=4xy^2=4x gives dy/dx=2/y=12dy/dx=2/y=\tfrac12 at P(4,4)P(4,4), so the normal has slope −2-2: y−4=−2(x−4)y-4=-2(x-4), i.e. x=6−y2x=6-\tfrac y2. Substituting into x=y2/4x=y^2/4 gives y2+2y−24=0y^2+2y-24=0, so y=4y=4 (point PP) or y=−6y=-6, giving Q(9,−6)Q(9,-6). Integrate with respect to yy, because the chord lies to the right of the parabola between these ordinates: A=∫−64[(6−y2)−y24]dy=[6y−y24−y312]−64=(24−4−163)−(−36−9+18)=443+27=1253A=\int_{-6}^{4}\left[\left(6-\tfrac y2\right)-\tfrac{y^2}{4}\right]dy=\left[6y-\tfrac{y^2}{4}-\tfrac{y^3}{12}\right]_{-6}^{4}=\left(24-4-\tfrac{16}{3}\right)-\left(-36-9+18\right)=\tfrac{44}{3}+27=\tfrac{125}{3}. The value 125/6125/6 is half the correct area, and 64/364/3 and 100/3100/3 are distractors.
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Original question written in the JEE Advanced 2016 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake