设\(P(x,y)\)为椭圆\(\dfrac{x^2}{16}+\dfrac{y^2}{12}=1\)在第一象限上的点,则\(\dfrac{x}{4-x}+\dfrac{3y}{6-y}\)的最小值为\(\underline{\qquad\qquad}\).
解析:
由题设$$P(4\cos\theta,2\sqrt{3}\sin\theta),\theta\in\left(0,\dfrac{\pi}{2}\right),$$ 则
\[ \begin{split}
&\dfrac{4}{4-x}+\dfrac{18}{6-y}-4\\
=&\dfrac{1}{1-\cos\theta}+\dfrac{3\sqrt{3}}{\sqrt{3}-\sin\theta}-4\\
=&\dfrac{1}{1-\cos \theta}+\dfrac{1}{1-\dfrac{\sin\theta}{\sqrt{3}}}+\dfrac{1}{1-\dfrac{\sin\theta}{\sqrt{3}}}+\dfrac{1}{1-\dfrac{\sin\theta}{\sqrt{3}}}-4\\
\geqslant& \dfrac{\left(1+1+1+1\right)^2}{\left(1-\cos\theta\right)+\left(1-\dfrac{\sin\theta}{\sqrt{3}}\right)+\left(1-\dfrac{\sin\theta}{\sqrt{3}}\right)+\left(1-\dfrac{\sin\theta}{\sqrt{3}}\right)}-4\\
=&\dfrac{16}{4-\cos\theta-\sqrt{3}\sin\theta}-4\\
\geqslant& 4.
\end{split}
\]
两处不等号的取等条件均为\(\theta=\dfrac{\pi}{3}\).因此当且仅当\(P\)的坐标为\(\left( 2,3\right)\)时,所求表达式取得最小值\(4\).
