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JEE Advanced 2025 Paper 2 · Mathematics · Conic Sections · medium

The normal to the parabola y^2=4x at the point (4,4) meets the parabola again at a point Q. Let Δ1 be the area of the region bounded by this normal…

Single correct option

The normal to the parabola y2=4xy^2=4x at the point (4,4)(4,4) meets the parabola again at a point QQ. Let Δ1\Delta_1 be the area of the region bounded by this normal and the parabola, and Δ2\Delta_2 the area of the region bounded by the parabola and its latus rectum. Then Δ1/Δ2\Delta_1/\Delta_2 equals
  1. A. 1258\dfrac{125}{8}
  2. B. 12516\dfrac{125}{16}
  3. C. 12512\dfrac{125}{12}
  4. D. 254\dfrac{25}{4}
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Answer: (A) 1258\dfrac{125}{8}

At (4,4)(4,4) the tangent slope is 2/y=122/y=\tfrac12, so the normal has slope −2-2: y=12−2xy=12-2x, i.e. x=12−y2x=\frac{12-y}{2}. Substituting in y2=4xy^2=4x gives y2+2y−24=0y^2+2y-24=0, so y=4y=4 or y=−6y=-6; Q=(9,−6)Q=(9,-6). Area between the chord and parabola (integrating in yy): Δ1=∫−64(12−y2−y24)dy=1253\Delta_1=\int_{-6}^{4}\left(\frac{12-y}{2}-\frac{y^2}{4}\right)dy=\frac{125}{3}. The latus rectum is x=1x=1 with ∣y∣≤2|y|\le2, so Δ2=∫−22(1−y24)dy=83\Delta_2=\int_{-2}^{2}\left(1-\frac{y^2}{4}\right)dy=\frac{8}{3}. Ratio =1258=\frac{125}{8}. Wrong options arise from halving areas or using a wrong second intersection.
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Original question written in the JEE Advanced 2025 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