93學年度指定科目考試數學(甲)非選擇題詳解

1. 二元一次方程組不只一組解 $\Rightarrow $相依方程組 $\Rightarrow $ $\displaystyle{\frac{1+cos\ \theta }{-1}}=\displaystyle{\frac{-1}{1+sin\ \theta }}$ $\Rightarrow $ $\left( 1+cos\theta \right)\left( 1+sin\theta \right)=1$ $\Rightarrow $ $1+sin\theta +cos\theta +sin\theta cos\theta =1$
   令$sin\theta +cos\theta =t$，由三角函數疊合可知$-\sqrt{2}\le t\le \sqrt{2}$，將$t$代入$1+cos\theta +sin\theta +sin\theta cos\theta =1$ $\Rightarrow $ $t+\displaystyle{\frac{\left( {{t}^{2}}-1 \right)}{2}}=0$ $\Rightarrow $ ${{t}^{2}}+2t-1=0$ $\Rightarrow $ $t=-1\pm \sqrt{2}$，但$-1-\sqrt{2}$不合($\because -\sqrt{2}\le t\le \sqrt{2}$) $\Rightarrow $ $t=-1+\sqrt{2}$
   
   
2. 令$P\left( 2，0 \right)$、$F\left( 1，2 \right)$，$L$：$kx+y+1=0$，如圖：
   
   $\Rightarrow $ $d\left( P，L \right)=d\left( P，F \right)$ $\Rightarrow $ $\sqrt{5}=\displaystyle{\frac{\left| 2k+2 \right|}{\sqrt{{{k}^{2}}+1}}}$ $\Rightarrow $ $5{{k}^{2}}+5=4{{k}^{2}}+4k+1$ $\Rightarrow $ ${{k}^{2}}-4k+4=0$ $\Rightarrow $ $k=2$ $\Rightarrow $準線：$2x+y+1=0$、對稱軸$x-2y+3=0$ $\Rightarrow $準點$Q\left( -1，1 \right)$ $\Rightarrow $頂點$=\displaystyle{\frac{F+Q}{2}}=\left( 0，\displaystyle{\frac{3}{2}} \right)$