Tangent lines are drawn to the ellipse (x^2/(25)+(y^2/(16)=1 at points of the first quadrant. Each tangent cuts off a segment between the coordinate…

A tangent at $(5\cos\theta,4\sin\theta)$ is $\frac{x\cos\theta}{5}+\frac{y\sin\theta}{4}=1$, with intercepts $\frac{5}{\cos\theta}$ and $\frac{4}{\sin\theta}$ on the axes. The squared length of the intercepted segment is $L^{2}=25\sec^{2}\theta+16\csc^{2}\theta=41+25\tan^{2}\theta+16\cot^{2}\theta$. By AM-GM, $25\tan^{2}\theta+16\cot^{2}\theta\ge2\sqrt{25\cdot16}=40$, with equality when $\tan^{4}\theta=\frac{16}{25}$, which occurs for an acute $\theta$. So $L^{2}_{\min}=41+40=81$, i.e. $L_{\min}=a+b=9$. A numerical scan over $\theta$ gives a minimum of $81.000$, confirming the result. The minimisation could also be done with calculus by differentiating $L^{2}$ with respect to $\theta$, which leads to the same critical point $\tan^{2}\theta=\frac45$. The general result is that the shortest intercepted tangent has length $a+b$, but the problem as set requires recognising the intercept form of the tangent and then applying AM-GM, rather than quoting a formula.