Let a variable line passing through a fixed point P in the first quadrant cuts the positive coordinate axes at points A and B respectively. If the area of Δ O A B is minimum, then O P is

Step-by-step solution

Let, the foot of perpendiculars from P on O A & O B are C & D respectively Let, ∠ P A C = θ = ∠ B P D Now, C A = k cot ⁡ θ & B D = h tan ⁡ θ Area of Δ O A B = area of Δ P C A + area of Δ B P D + area of the rectangle P D O C ⇒ Area of Δ O A B = 1 2 k 2 cot ⁡ θ + 1 2 h 2 tan ⁡ θ + h k = 1 2 k cot ⁡ θ - h tan ⁡ θ 2 + 2 h k ⇒ The minimum area of Δ O A B is 2 h k when k cot ⁡ θ = h tan