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 ______

The equation of the curve is:

$$9x^2 + 16y^2 = 144$$

Dividing both sides by $$144$$:

$$\frac{x^2}{16} + \frac{y^2}{9} = 1$$

This is a standard ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, where:

• $$a^2 = 16 \implies a = 4$$
• $$b^2 = 9 \implies b = 3$$

Step 1: Equation of a Parametric Tangent

Any parametric point on this ellipse can be represented as $$(a \cos\theta, b \sin\theta) = (4 \cos\theta, 3 \sin\theta)$$.

The equation of the tangent to the ellipse at this parametric point is given by:

$$\frac{x \cos\theta}{a} + \frac{y \sin\theta}{b} = 1$$

Substituting $$a = 4$$ and $$b = 3$$:

$$\frac{x \cos\theta}{4} + \frac{y \sin\theta}{3} = 1$$

Step 2: Find the Intercepts $$A$$ and $$B$$

The tangent line intersects the coordinate axes at points $$A$$ (on the x-axis) and $$B$$ (on the y-axis).

• For point $$A$$, set $$y = 0$$:$$\frac{x \cos\theta}{4} = 1 \implies x = \frac{4}{\cos\theta} \implies A = \left(\frac{4}{\cos\theta}, 0\right)$$
• For point $$B$$, set $$x = 0$$:$$\frac{y \sin\theta}{3} = 1 \implies y = \frac{3}{\sin\theta} \implies B = \left(0, \frac{3}{\sin\theta}\right)$$

Step 3: Find the Length of Segment $$AB$$

Using the distance formula, the length $$L$$ of the segment $$AB$$ is:

$$L = \sqrt{\left(\frac{4}{\cos\theta} - 0\right)^2 + \left(0 - \frac{3}{\sin\theta}\right)^2}$$  $$L = \sqrt{\frac{16}{\cos^2\theta} + \frac{9}{\sin^2\theta}}$$

To minimize $$L$$, we can minimize $$L^2$$:

$$f(\theta) = L^2 = 16\sec^2\theta + 9\csc^2\theta$$

Step 4: Minimize $$L^2$$

Using the standard identity $$\sec^2\theta = 1 + \tan^2\theta$$ and $$\csc^2\theta = 1 + \cot^2\theta$$:

$$f(\theta) = 16(1 + \tan^2\theta) + 9(1 + \cot^2\theta)$$  $$f(\theta) = 25 + 16\tan^2\theta + 9\cot^2\theta$$

Using the AM-GM Inequality ($$A \geq G$$) on the variable terms $$16\tan^2\theta$$ and $$9\cot^2\theta$$:

$$\frac{16\tan^2\theta + 9\cot^2\theta}{2} \geq \sqrt{16\tan^2\theta \cdot 9\cot^2\theta}$$  $$16\tan^2\theta + 9\cot^2\theta \geq 2 \cdot \sqrt{144 \cdot (\tan^2\theta \cdot \cot^2\theta)}$$

Since $$\tan^2\theta \cdot \cot^2\theta = 1$$:

$$16\tan^2\theta + 9\cot^2\theta \geq 2 \cdot 12 = 24$$

Substituting this back into our expression for $$f(\theta)$$:

$$f(\theta)_{\text{min}} = 25 + 24 = 49$$

Since $$L^2 \geq 49$$, the minimum length $$L$$ is:

$$L_{\text{min}} = \sqrt{49} = 7$$

Answer:

7