A vertical line passing through the point h , 0 intersects the ellipse x 2 4 + y 2 3 = 1 at the points P and Q . Let the tangents to the ellipse at P and Q meet at the point R . If Δ ( h ) = area of triangle PQR , Δ 1 = max 1 2 ≤ h ≤ 1 Δ h and Δ 2 = min 1 2 ≤ h ≤ 1 Δ ( h ) , then 8 5 Δ 1 - 8 Δ 2 =

Step-by-step solution

Point of intersection of tangents at P and Q is R 2 sec θ , 0 Area of Δ PQR = 1 2 · 2 3 sin θ · 2 sec θ - 2 cos θ ⇒ Δ = 2 3 · sin 3 θ cos θ ; where cos θ ∈ 1 4 , 1 2 Now d Δ d θ = 2 3 cos θ · 3 sin 2 θ cos θ - sin 3 θ - sin θ cos 2 θ > 0 As θ increases, Δ increases ⇒ when cos θ decreases, Δ increases ∴ Δ min occurs at cos θ = 1 / 2 , Therefore Δ 2 = 2 3 · 1 - 1 / 4 3 / 2 1 / 2 = 4 3 · 3 3 8 = 3 6 8 Δ max occurs at cos θ = 1 / 4 , Therefore Δ 1 = 2 3 · 1 - 1 / 1 6 3 / 2 1 / 4 = 8 3 · 1 5 . 1 5 4.4.4 = 2 3 . 1 5 . 3