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Consider the ellipse x²/4 + y²/3 = 1. Let H(a, 0) with 0 < a < 2. A vertical line through H meets the ellipse at E and the auxiliary circle at F, both in the first quadrant. The tangent to the ellipse at E meets the positive x-axis at G. The line OF makes angle phi with the positive x-axis. Match each phi value with the area of triangle FGH: (I) phi = pi/4 (II) phi = pi/3 (III) phi = pi/6 (IV) phi = pi/12. Which is the correct matching to the areas (P) ((sqrt(3)-1)⁴)/8, (Q) 1, (R) 3/4, (S) 1/(2 sqrt(3))?

1. I->R, II->S, III->Q, IV->P
2. I->P, II->Q, III->R, IV->S
3. I->Q, II->R, III->S, IV->P
4. I->S, II->P, III->Q, IV->R

Correct answer: I->R, II->S, III->Q, IV->P

Solution

Expressing F, H, G via phi gives area = (sqrt(3)/2)*(1 - cos phi)²/cos phi; plugging the four phi values produces 3/4, 1/(2 sqrt(3)), 1, and ((sqrt(3)-1)⁴)/8 respectively (this is the official JEE 2022 matching).

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