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JEE Advanced 2016 Paper 2 · Mathematics · Conic Sections · hard

JEE Advanced 2016 Paper 2 Mathematics: Paragraph 1

Consider the parabola y2=4xy^{2}=4x with focus FF. A focal chord PQPQ of the parabola, with PP in the first quadrant, satisfies FP=4 FQFP=4\,FQ. The tangents to the parabola at PP and QQ meet at the point TT. Recall that the points of the parabola may be written as (t2,2t)(t^{2},2t), that the tangent at (t2,2t)(t^{2},2t) is ty=x+t2ty=x+t^{2}, and that the ends of a focal chord have parameters tt and −1/t-1/t.

Question 1

Single correct option

The area of triangle TPQTPQ is
  1. A. 1258\dfrac{125}{8}
  2. B. 62564\dfrac{625}{64}
  3. C. 12516\dfrac{125}{16}
  4. D. 12532\dfrac{125}{32}
Show answer and solution

Answer: (C) 12516\dfrac{125}{16}

With F=(1,0)F=(1,0), FP=1+t2FP=1+t^{2} and FQ=1+1t2=1+t2t2FQ=1+\frac1{t^{2}}=\frac{1+t^{2}}{t^{2}} (distance to the directrix). So FPFQ=t2=4\frac{FP}{FQ}=t^{2}=4, giving t=2t=2 (PP in the first quadrant). Then P=(4,4)P=(4,4), Q=(14,−1)Q=\left(\frac14,-1\right) and the tangents 2y=x+42y=x+4 and −12y=x+14-\frac12y=x+\frac14, i.e. y=−2x−12y=-2x-\frac12, meet at T=(−1,t−1t)=(−1,32)T=(-1,t-\frac1t)=\left(-1,\frac32\right). The slopes 12\frac12 and −2-2 have product −1-1, so ∠PTQ=90∘\angle PTQ=90^{\circ}. Then TP=25+254=552TP=\sqrt{25+\frac{25}{4}}=\frac{5\sqrt5}{2} and TQ=2516+254=554TQ=\sqrt{\frac{25}{16}+\frac{25}{4}}=\frac{5\sqrt5}{4}. Area =12⋅552⋅554=12516=\frac12\cdot\frac{5\sqrt5}{2}\cdot\frac{5\sqrt5}{4}=\frac{125}{16}, which matches a determinant computation.

Question 2

Single correct option

The radius of the circle passing through PP, QQ and TT is
  1. A. 2516\dfrac{25}{16}
  2. B. 258\dfrac{25}{8}
  3. C. 254\dfrac{25}{4}
  4. D. 52\dfrac{5}{2}
Show answer and solution

Answer: (B) 258\dfrac{25}{8}

From the data, P=(4,4)P=(4,4) and Q=(14,−1)Q=\left(\frac14,-1\right), so the focal chord has length PQ=(4−14)2+(4+1)2=22516+25=62516=254PQ=\sqrt{\left(4-\frac14\right)^{2}+(4+1)^{2}}=\sqrt{\frac{225}{16}+25}=\sqrt{\frac{625}{16}}=\frac{25}{4}; equally FP+FQ=5+54=254FP+FQ=5+\frac54=\frac{25}{4}. The tangents at the ends of a focal chord are perpendicular, and T=(−1,32)T=\left(-1,\frac32\right) gives TP⃗⋅TQ⃗=(5)(54)+(52)(−52)=0\vec{TP}\cdot\vec{TQ}=(5)(\tfrac54)+\left(\tfrac52\right)(-\tfrac52)=0. Hence ∠PTQ=90∘\angle PTQ=90^{\circ} and, by Thales' theorem, PQPQ is a diameter of the circumcircle of TPQTPQ. The radius is 12⋅254=258\frac12\cdot\frac{25}{4}=\frac{25}{8}. Option (C) is the diameter and option (A) is half the correct radius.
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Original question written in the JEE Advanced 2016 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake