Let a tangent to the curve $9x^{2} + 16y^{2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is ____.
Step-by-Step Solution
Key Concept: Write the ellipse as $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Tangent at parametric point $P(4\cos\theta, 3\sin\theta)$ is $\frac{x\cos\theta}{4} + \frac{y\sin\theta}{3} = 1$. Intercepts give $A = (4\sec\theta, 0)$, $B = (0, 3\csc\theta)$. Minimize $|AB|^2 = 16\sec^2\theta + 9\csc^2\theta$.
Tangent at $P(4\cos\theta, 3\sin\theta)$: $\frac{x\cos\theta}{4} + \frac{y\sin\theta}{3} = 1$. So $A = (4\sec\theta, 0)$, $B = (0, 3\csc\theta)$. $|AB|^2 = 16\sec^2\theta + 9\csc^2\theta = 25 + 16\tan^2\theta + 9\cot^2\theta \geq 25 + 2\sqrt{144} = 49$. Minimum $|AB| = 7$.
Correct Answer: 7