Let a tangent to the curve 9x² + 16y² = 144 intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is ________

Solution

Maths sir · 4 steps

Use the parametric tangent, write AB² in terms of θ, and let AM-GM find the minimum.

1. Standard form. Dividing by 144 gives x²/16 + y²/9 = 1, so a = 4 and b = 3. Take the point (4 cos θ, 3 sin θ), with θ not a multiple of π/2 so that the tangent meets both axes.
2. Tangent and intercepts. The tangent there is (x cos θ)/4 + (y sin θ)/3 = 1. It cuts the axes at A(4/(cos θ), 0) and B(0, 3/(sin θ)).

   AB² = 16 sec² θ + 9 csc² θ = 25 + 16 tan² θ + 9 cot² θ

3. Apply AM-GM. 16 tan² θ + 9 cot² θ ≥ 2√(16 × 9) = 24, because tan² θ cot² θ = 1. Equality holds when tan² θ = 3/4, so the minimum is reached. That gives AB² ≥ 25 + 24 = 49.
4. So. The minimum length of AB is √49 = 7.

The common slip: 49. Stopping at AB² gives 49. The question asks for the length, which is √49 = 7.

The idea to keep. For the ellipse x²/a² + y²/b² = 1, the intercept cut off by a tangent between the axes has minimum length a + b. Here that is 4 + 3 = 7.

Answer from the NTA final answer key. Solution by PrepWiser.