Tangents are drawn to the ellipse , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
The ellipse is x²/2 + y²/1 = 1, so a² = 2, b² = 1. A tangent at point (√2 cosθ, sinθ) is:
x·cosθ/√2 + y·sinθ = 1
X-intercept (set y=0): x = √2/cosθ
Y-intercept (set x=0): y = 1/sinθ
Midpoint (h, k) of the intercept segment between the axes:
h = (1/2)(√2/cosθ) → cosθ = √2/(2h) = 1/(√2·h)
k = (1/2)(1/sinθ) → sinθ = 1/(2k)
Using cos²θ + sin²θ = 1:
1/(2h²) + 1/(4k²) = 1
Replacing h,k with x,y: 1/(2x²) + 1/(4y²) = 1, which is option 1.
x·cosθ/√2 + y·sinθ = 1
X-intercept (set y=0): x = √2/cosθ
Y-intercept (set x=0): y = 1/sinθ
Midpoint (h, k) of the intercept segment between the axes:
h = (1/2)(√2/cosθ) → cosθ = √2/(2h) = 1/(√2·h)
k = (1/2)(1/sinθ) → sinθ = 1/(2k)
Using cos²θ + sin²θ = 1:
1/(2h²) + 1/(4k²) = 1
Replacing h,k with x,y: 1/(2x²) + 1/(4y²) = 1, which is option 1.
[AIEEE 2004]