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JEE Main 2015

JEE Main 2015 Mathematics question on Conic Sections

The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latusrectum to the ellipse x29+y25=1\frac{x^2}{9} + \frac{y^2}{5} = 1 is
  1. A.
    274\frac{27}{4}
  2. B.
    1818
  3. C.
    272\frac{27}{2}
  4. D.
    2727
Show answer and solution

Correct option: D

Solution

For the ellipse x²/9 + y²/5 = 1, a² = 9, b² = 5, so c = √(9-5) = 2. The endpoints of the latus rectum are at (±2, ±5/3).

The tangent at point (x₀, y₀) on the ellipse is xx₀/9 + yy₀/5 = 1. At endpoint (2, 5/3): 2x/9 + y/3 = 1, giving x-intercept (9/2, 0) and y-intercept (0, 3). By symmetry, the four tangents at the four latus rectum endpoints form a rhombus with vertices at (±9/2, 0) and (0, ±3).

Area of rhombus = (1/2) × d₁ × d₂ = (1/2) × 9 × 6 = 27 sq units.

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