A vertical line passing through the point (h,0) intersects the ellipse x²/4 + y²/3 = 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 the triangle PQR, Δ₁ = max₁/₂≤h≤₁ Δ(h) and Δ₂ = min₁/₂≤h≤₁ Δ(h), then 8√5Δ₁/Δ₂ = ?

Solution:
x²/4 + y²/3 = 1
y = √(3/2)√(4 - h²)
at x = h
Let R = (x₁, 0)
PQ is a chord of contact, so xx₁/4 = 1 ⇒ x = 4/x₁ which is equation of PQ, x = h
so, 4/x₁ = h ⇒ x₁ = 4/h
Δ(h) = ar(ΔPQR) = 1/2 × PQ × RT
1/2 × 2√(3/2)√(4 - h²) × (x₁ - h) = √(3/2)h(4 - h²)^3/2
Δ'(h) = -√3(4 + 2h²)/2h²√(4 - h²)
which is always decreasing.
so Δ₁ = maximum of Δ(h) = (4√5)/8 at h = 1/2
Δ₂ = minimum of Δ(h) = 9/2 at h = 1
so 8√5Δ₁/Δ₂ = 8√5 × (4√5)/8 / (9/2) = 4√5²/9/2 = 4(5)/9/2 = 40/9
Hence, the given expression is not equal to any of the given options. There might be an error in the problem statement or the given options.