The minimum length of the intercept between the coordinate axes made by a tangent of the ellipse $\frac{{\mathrm{x}}^2}{64}+\frac{{\mathrm{y}}^2}{4}=1$

Given ellipse $\frac{{\mathrm{x}}^2}{64}+\frac{{\mathrm{y}}^2}{4}=1$

where ${\mathrm{a}}^2=64, {\mathrm{b}}^2=4$

Equation of tangent at $\mathrm{P}\left(\mathrm{\theta }\right)$ is $\frac{\mathrm{x}}{8}\mathrm{cos\theta }+\frac{\mathrm{y}}{2}\mathrm{sin\theta }=1$

It cuts x-axis at $\mathrm{A}=\left(\frac{8}{\mathrm{cos\theta }}, 0\right)$ and y-axis at $\mathrm{B}=\left(0, \frac{2}{\mathrm{sin\theta }}\right)$

Now  $\mathrm{AB}=\sqrt{\frac{64}{{\cos }^2\mathrm{\theta }}+\frac{4}{{\sin }^2\mathrm{\theta }}}$

               $=2\sqrt{16{\sec }^2\mathrm{\theta }+{\mathrm{cosec}}^2\mathrm{\theta }}$

Now minimum value of $\mathrm{AB}=2(4+1)=10$