回復 1# EZWrookie 的帖子
第3題
直線\(y=x+1\)與圓O：\({{x}^{2}}+{{y}^{2}}=10\)交於\({{F}_{1}},{{F}_{3}}\)，與\({{x}^{2}}=4y\)交於\({{F}_{2}}\left( 2-2\sqrt{2},3-2\sqrt{2} \right),{{F}_{4}}\left( 2+2\sqrt{2},3+2\sqrt{2} \right)\)
而\(F\left( 0,1 \right)\)在\(\overline{{{F}_{2}}{{F}_{3}}}\)上，\({{x}^{2}}=4y\)的準線是\(y=-1\)
先用根與係數算出\(\overline{{{F}_{1}}{{F}_{3}}}=\sqrt{38}\)，作\(\overline{OP}\)垂直\(y=x+1\)於P
\(\begin{align}
& \overline{OP}=\overline{PF}=\frac{\sqrt{2}}{2},\overline{F{{F}_{1}}}=\frac{\sqrt{38}}{2}+\frac{\sqrt{2}}{2},\overline{F{{F}_{3}}}=\frac{\sqrt{38}}{2}-\frac{\sqrt{2}}{2} \\
& \overline{F{{F}_{4}}}=4+2\sqrt{2},\overline{F{{F}_{2}}}=4-2\sqrt{2} \\
& \overline{{{F}_{1}}{{F}_{2}}}+\overline{{{F}_{3}}{{F}_{4}}}=\overline{F{{F}_{1}}}-\overline{F{{F}_{2}}}+\overline{F{{F}_{4}}}-\overline{F{{F}_{3}}}=\left( \overline{F{{F}_{1}}}-\overline{F{{F}_{3}}} \right)+\left( \overline{F{{F}_{4}}}-\overline{F{{F}_{2}}} \right)=5\sqrt{2} \\
\end{align}\)
