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ExamsJEE MainMaths

Two ellipses E1 and E2 are centred at the origin; the major axis of E1 lies along the x-axis and that of E2 along the y-axis. The circle S: x² + (y - 1)² = 2 and the line x + y = 3 touch S, E1 and E2 at points P, Q and R respectively. Given PQ = PR = 2*sqrt(2)/3, and letting e1, e2 be the eccentricities of E1 and E2, which expression(s) is/are correct?

  1. e1² + e2² = 43/40
  2. e1*e2 = sqrt(7)/(2*sqrt(10))
  3. |e1² - e2²| = 5/8
  4. e1*e2 = sqrt(3)/4

Correct answer: e1² + e2² = 43/40

Solution

Solving the geometry (P = (2,1), Q and R found from PQ = PR = 2*sqrt(2)/3) and the tangency conditions yields a1² = 8, b1² =... such that e1² + e2² = 43/40 and e1*e2 = sqrt(7)/(2*sqrt(10)).

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