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JEE Main 2019 · 10 April · Shift I

JEE Main 2019 Mathematics question on Conic Sections

If the line x−2y=12x - 2y = 12 is tangent to the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 at the point (3,−92)\left(3, \frac{-9}{2}\right), then the length of the latus rectum of the ellipse is
  1. A.
    838\sqrt{3}
  2. B.
    99
  3. C.
    55
  4. D.
    12212\sqrt{2}
Show answer and solution

Correct option: B

Solution

The tangent to the ellipse at point (3, -9/2) is given by (3x/a²) + ((-9/2)y/b²) = 1, which must match x - 2y = 12, i.e., x/12 - y/6 = 1.

Comparing: 3/a² = 1/12 → a² = 36, and (-9/2)/b² = -1/6 → b² = 27.

Length of latus rectum = 2b²/a = 2(27)/6 = 9.

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