The normal at a point on the ellipse meets the -axis at . If is the mid-point of the line segment , then the locus of intersects the latusrectum of the given ellipse at the points
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
The ellipse is x²/16 + y²/4 = 1, so a² = 16, b² = 4, giving c = √12 = 2√3. The latus rectum is at x = ±2√3.
Parametrize P = (4cosθ, 2sinθ). The normal at P to x²/16 + y²/4 = 1 has equation:
4x/cosθ - 2y/sinθ = 12 (using a²x/x₀ - b²y/y₀ = a² - b²).
At Q (where y = 0): 4x/cosθ = 12, so Q = (3cosθ, 0).
M is the midpoint of PQ:
h = (4cosθ + 3cosθ)/2 = (7cosθ)/2
k = (2sinθ + 0)/2 = sinθ
So cosθ = 2h/7 and sinθ = k. Using cos²θ + sin²θ = 1:
4h²/49 + k² = 1, which is the locus: x²/(49/4) + y²/1 = 1.
The latus rectum of the original ellipse is at x = ±2√3. Substituting:
4(12)/49 + y² = 1 → y² = 1 - 48/49 = 1/49 → y = ±1/7.
The intersection points are (±2√3, ±1/7), which is option 3.
Parametrize P = (4cosθ, 2sinθ). The normal at P to x²/16 + y²/4 = 1 has equation:
4x/cosθ - 2y/sinθ = 12 (using a²x/x₀ - b²y/y₀ = a² - b²).
At Q (where y = 0): 4x/cosθ = 12, so Q = (3cosθ, 0).
M is the midpoint of PQ:
h = (4cosθ + 3cosθ)/2 = (7cosθ)/2
k = (2sinθ + 0)/2 = sinθ
So cosθ = 2h/7 and sinθ = k. Using cos²θ + sin²θ = 1:
4h²/49 + k² = 1, which is the locus: x²/(49/4) + y²/1 = 1.
The latus rectum of the original ellipse is at x = ±2√3. Substituting:
4(12)/49 + y² = 1 → y² = 1 - 48/49 = 1/49 → y = ±1/7.
The intersection points are (±2√3, ±1/7), which is option 3.
[AIEEE 2009]