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JEE Advanced 2019 Paper 1 · Mathematics · Conic Sections · hard

(x^2/(16)+(y^2/(7)=1 with foci F1(3,0) and F2(-3,0).

One or more options correct

Consider the ellipse E: x216+y27=1E:\ \dfrac{x^2}{16}+\dfrac{y^2}{7}=1 with foci F1(3,0)F_1(3,0) and F2(−3,0)F_2(-3,0). Which of the following statements is/are correct?
  1. A. The four tangents drawn at the end points of the two latera recta of EE enclose a quadrilateral of area 1283\dfrac{128}{3}
  2. B. For every tangent to EE, the product of the perpendicular distances of F1F_1 and F2F_2 from it is 77
  3. C. The locus of the foot of the perpendicular from F1F_1 on a variable tangent to EE is the circle x2+y2=7x^2+y^2=7
  4. D. The length of a latus rectum of EE is 74\dfrac74
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Answer: A, B

Here a2=16a^2=16, b2=7b^2=7, c=3c=3, e=34e=\frac34. The latus rectum has length 2b2a=72\frac{2b^2}{a}=\frac72, so (D) is false. The ends of the latus rectum through F1F_1 are (3,±74)(3,\pm\frac74); the tangent at (3,74)(3,\frac74) is 3x16+y4=1\frac{3x}{16}+\frac y4=1, with intercepts 163\frac{16}3 and 44. The four tangents form a rhombus whose diagonals are 2⋅1632\cdot\frac{16}3 and 2⋅42\cdot4, so the area is 12⋅323⋅8=1283\frac12\cdot\frac{32}3\cdot8=\frac{128}3 (A true). For the tangent xcos⁡θ4+ysin⁡θ7=1\frac{x\cos\theta}4+\frac{y\sin\theta}{\sqrt7}=1, the product of the distances from (±3,0)(\pm3,0) is ∣1−9cos⁡2θ16∣cos⁡2θ16+sin⁡2θ7=7\frac{|1-\frac{9\cos^2\theta}{16}|}{\frac{\cos^2\theta}{16}+\frac{\sin^2\theta}{7}}=7 (B true). The foot of the perpendicular from a focus lies on the auxiliary circle x2+y2=16x^2+y^2=16, not x2+y2=7x^2+y^2=7 (C false).
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