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楼主 |
发表于 2014-4-28 22:28:51
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若8#所列的关系式正确,我们可以得到双椭圆计算公式
外切于椭圆\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\),内接于椭圆\(\frac{(x-x_0)^2}{m^2}+\frac{(y-y_1)}{n^2}=1\)
\(n=3\)时
\((a^2n^2+b^2m^2+m^2n^2-m^2y_0^2-n^2x_0^2)^2+4(-a^2b^2-a^2n^2+a^2y_0^2-b^2m^2+b^2x_0^2)n^2m^2=0\)
\(n=4\)时
\(-(a^2n^2+b^2m^2+m^2n^2-m^2y_0^2-n^2x_0^2)^3-4n^2m^2(-a^2b^2-a^2n^2+a^2y_0^2-b^2m^2+b^2x_0^2)(a^2n^2+b^2m^2+m^2n^2-m^2y_0^2-n^2x_0^2)-8a^2b^2n^4m^4=0\)
\(n=5\)时
\((a^2n^2+b^2m^2+m^2n^2-m^2y_0^2-n^2x_0^2)^6+12n^2m^2(a^2n^2+b^2m^2+m^2n^2-m^2y_0^2-n^2x_0^2)^4(-a^2b^2-a^2n^2+a^2y_0^2-b^2m^2+b^2x_0^2)-32n^4m^4(a^2n^2+b^2m^2+m^2n^2-m^2y_0^2-n^2x_0^2)^3a^2b^2+48n^4m^4(a^2n^2+b^2m^2+m^2n^2-m^2y_0^2-n^2x_0^2)^2(-a^2b^2-a^2n^2+a^2y_0^2-b^2m^2+b^2x_0^2)^2-128(a^2n^2+b^2m^2+m^2n^2-m^2y_0^2-n^2x_0^2)(-a^2b^2-a^2n^2+a^2y_0^2-b^2m^2+b^2x_0^2)a^2b^2m^6n^6+64m^6n^6(-a^2b^2-a^2n^2+a^2y_0^2-b^2m^2+b^2x_0^2)^3-256a^4b^4n^8m^8=0\)
若\(x_0=0,y_0=0\),则有下面结论:
\(n=3\)时
\((a^2n^2+b^2m^2+m^2n^2)^2+4(-a^2b^2-a^2n^2-b^2m^2)n^2m^2=0\)
\(n=4\)时
\(-(a^2n^2+b^2m^2+m^2n^2)^3-4n^2m^2(-a^2b^2-a^2n^2-b^2m^2)(a^2n^2+b^2m^2+m^2n^2)-8a^2b^2n^4m^4=0\)
\(n=5\)时
\((a^2n^2+b^2m^2+m^2n^2)^6+12n^2m^2(a^2n^2+b^2m^2+m^2n^2)^4(-a^2b^2-a^2n^2-b^2m^2)-32n^4m^4(a^2n^2+b^2m^2+m^2n^2)^3a^2b^2+48n^4m^4(a^2n^2+b^2m^2+m^2n^2)^2(-a^2b^2-a^2n^2-b^2m^2)^2-128(a^2n^2+b^2m^2+m^2n^2)(-a^2b^2-a^2n^2-b^2m^2)a^2b^2m^6n^6+64m^6n^6(-a^2b^2-a^2n^2-b^2m^2)^3-256a^4b^4n^8m^8=0\)
谁有兴趣验证一下上面的公式是否正确?
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