三正交射线交椭球而得平面的极点轨迹
本帖最后由 葡萄糖 于 2020-6-7 19:31 编辑当三正交射线的端点在椭球\(\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}+\tfrac{z^2}{c^2}=1\)中心\(O(0,0,0)\)时，所确定的平面为
\[ x_{0} x+y_{0} y+z_{0}z=\dfrac{1}{\tfrac{1}{a^2}+\tfrac{1}{b^2}+\tfrac{1}{c^2}} \]
其中\({x_0}^2+{y_0}^2+{z_0}^2=\dfrac{1}{\tfrac{1}{a^2}+\tfrac{1}{b^2}+\tfrac{1}{c^2}}\)
极点的轨迹为
\[ \dfrac{x^2}{a^4\left(\tfrac{1}{a^2}+\tfrac{1}{b^2}+\tfrac{1}{c^2}\right)}+\dfrac{y^2}{b^4\left(\tfrac{1}{a^2}+\tfrac{1}{b^2}+\tfrac{1}{c^2}\right)}+\dfrac{z^2}{c^4\left(\tfrac{1}{a^2}+\tfrac{1}{b^2}+\tfrac{1}{c^2}\right)}=1 \]
或写成\( \dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \)
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