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{qquadqquad}).
解析:
由题设$$P(4cos heta,2sqrt{3}sin heta), hetainleft(0,dfrac{pi}{2}
ight),$$ 则
[ egin{split}
&dfrac{4}{4-x}+dfrac{18}{6-y}-4\
=&dfrac{1}{1-cos heta}+dfrac{3sqrt{3}}{sqrt{3}-sin heta}-4\
=&dfrac{1}{1-cos heta}+dfrac{1}{1-dfrac{sin heta}{sqrt{3}}}+dfrac{1}{1-dfrac{sin heta}{sqrt{3}}}+dfrac{1}{1-dfrac{sin heta}{sqrt{3}}}-4\
geqslant& dfrac{left(1+1+1+1
ight)^2}{left(1-cos heta
ight)+left(1-dfrac{sin heta}{sqrt{3}}
ight)+left(1-dfrac{sin heta}{sqrt{3}}
ight)+left(1-dfrac{sin heta}{sqrt{3}}
ight)}-4\
=&dfrac{16}{4-cos heta-sqrt{3}sin heta}-4\
geqslant& 4.
end{split}
]
两处不等号的取等条件均为( heta=dfrac{pi}{3}).因此当且仅当(P)的坐标为(left( 2,3 ight))时,所求表达式取得最小值(4).