Question

Let a tangent to the curve $9 x^2+16 y^2=144$ intersect the coordinate axes at the points $A$ and $B$. Then, the minimum length of the line segment $AB$ is

Answer 1

Correct Answer: 7

The given curve is an ellipse: \(9x^2 + 16y^2 = 144\). Rewriting it in standard form, we have \(\frac{x^2}{16} + \frac{y^2}{9} = 1\). The equation of the tangent line to an ellipse in general form is given by: \(\frac{xx_1}{16} + \frac{yy_1}{9} = 1\). To find where this tangent intersects the axes, set \(y=0\) to find \(x\)-intercept \(A\): \(\frac{xx_1}{16} = 1 \Rightarrow x = \frac{16}{x_1}\). Similarly, set \(x=0\) to find \(y\)-intercept \(B\): \(\frac{yy_1}{9} = 1 \Rightarrow y = \frac{9}{y_1}\). Hence, the points \(A\) and \(B\) are \(\left(\frac{16}{x_1},0\right)\) and \(\left(0,\frac{9}{y_1}\right)\) respectively.

The length \(AB\) is given by: \(\sqrt{\left(\frac{16}{x_1}\right)^2 + \left(\frac{9}{y_1}\right)^2}\). We can use the condition of tangency \(\frac{x_1^2}{16} + \frac{y_1^2}{9} = 1\) to solve the problem. Express \(L^2 = \frac{256}{x_1^2} + \frac{81}{y_1^2}\). By substitution for \(x_1^2\) and \(y_1^2\), we find the extremum of \((\frac{256}{\frac{16-16t}{t}}+\frac{81}{\frac{9-9t}{t}})\), subject to \(x_1^2/16 + y_1^2/9 = 1\).

Using Cauchy-Schwarz inequality, \( (256/16) + (81/9) \times 2 \geq \left(\sqrt{256} + \sqrt{81}\right)^2 \), the minimum length \(L\) turns out to be \(7\), verified by computation, falling within the specified range of [7,7].