https://mathmanstudio.com/categories/%E7%A9%BA%E9%96%93%E5%90%91%E9%87%8F/Recent content in 空間向量 on MathManStudio.comHugozh-TWWed, 01 Jul 2026 20:36:41 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a-%E5%A4%9A%E9%81%B88/Wed, 01 Jul 2026 20:36:41 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a-%E5%A4%9A%E9%81%B88/<p><img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2026/07/%E6%88%AA%E5%9C%96-2026-07-01-%E6%99%9A%E4%B8%8A7.33.00-1024x397.png"></p> <p>將原式配方整理後可得圓方程式如下：</p> <p>$$ （x-1）^2+y^2=4 $$</p> <p>此為圓心（1,0）半徑=2之圓方程式</p> <p>$$ (1)直線方程式x=3與此圓方程式之切線，交於1點，有兩相等實根。 $$</p> <p>（2）此圖形並未於原點有點對稱的現象。</p> <p>（3）正確。</p> <p>（4）解方程式判斷有無交點即可。</p> <p>$$ y=-x+4\quad代入（x-1）^2+y^2=4 $$ $$ 整理後可得x^2-5x+\frac{13}{2}=0 $$ $$ 判別式&lt;0\rightarrow無交點 $$</p> <p>(5)求極值當然用柯西不等式啦！</p> <p>$$ 令\vec{a}=(x-1,y)\quad\vec{b}=(1,-1) $$ $$ [(x-1)^2+y^2]\times[1^2+(-1)^2]\ge（x-1-y）^2 $$ $$ 1-2\sqrt{2}\le x-y\le1+2\sqrt{2} $$</p> <p>正解：（3）（5）</p>https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a%E5%96%AE%E9%81%B86/Sun, 28 Dec 2025 20:32:11 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a%E5%96%AE%E9%81%B86/<p><img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2025/10/%E6%88%AA%E5%9C%96-2025-10-27-%E6%99%9A%E4%B8%8A7.06.26-1024x774.png"> <img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2025/10/%E6%88%AA%E5%9C%96-2025-10-28-%E6%99%9A%E4%B8%8A8.00.37-1024x167.png"> <img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2025/12/%E5%96%AE%E9%81%B86-793x1024.jpg"></p> 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