The area of the quadrilateral formed by the tangents at the end points of latusrectum to the ellipse , is
- A.27/4 sq units
- B.9 sq units
- C.27/2 sq units
- D.27 sq units
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) (since b²/a = 5/3).
The tangent at point (x₀, y₀) on the ellipse is xx₀/9 + yy₀/5 = 1. At the end point (2, 5/3): 2x/9 + (5/3)y/5 = 1 → 2x/9 + y/3 = 1. This meets the axes at x = 9/2 (when y=0) and y = 3 (when x=0).
By symmetry, the four tangents at the four latus rectum endpoints form a rhombus with vertices on the axes at (±9/2, 0) and (0, ±3).
Area of rhombus = (1/2) × d₁ × d₂ = (1/2) × 9 × 6 = 27 sq units.
The tangent at point (x₀, y₀) on the ellipse is xx₀/9 + yy₀/5 = 1. At the end point (2, 5/3): 2x/9 + (5/3)y/5 = 1 → 2x/9 + y/3 = 1. This meets the axes at x = 9/2 (when y=0) and y = 3 (when x=0).
By symmetry, the four tangents at the four latus rectum endpoints form a rhombus with vertices on the axes at (±9/2, 0) and (0, ±3).
Area of rhombus = (1/2) × d₁ × d₂ = (1/2) × 9 × 6 = 27 sq units.
[AIEEE 2003]