本帖最后由 hbghlyj 于 2020-9-6 16:02 编辑
$y=x^3$与$(y+x-\alpha)^3=(y-\beta)^2$二阶切触，求$\alpha,\beta$
设切点为$(a,a^3)$,则$\led a^3&=\beta +s^3\\a&=s^2-\beta -s^3+\alpha\\3 a^2&=\frac{3 s^2}{2 s-3 s^2}\\6 a&=\frac{6 s}{2-6 s}\endled$
解得$(a,\alpha,\beta,s)=(0,0,0,0),\left(\frac{1}{6} \left(\sqrt{13}-1\right),\frac{1}{702} \left(181 \sqrt{13}-331\right),\frac{1}{169} \left(14 \sqrt{13}-39\right),\frac{1}{39} \left(13-\sqrt{13}\right)\right),\left(\frac{1}{6} \left(-\sqrt{13}-1\right),\frac{1}{702} \left(-181 \sqrt{13}-331\right),\frac{1}{169} \left(-14 \sqrt{13}-39\right),\frac{1}{39} \left(\sqrt{13}+13\right)\right)$
$y=x^3$与$(y+x-\alpha)^3=-1+(y-\beta)^2$二阶切触，求$\alpha,\beta$
设切点为$(a,a^3)$,则$\led (a^3+a-\alpha)^3&=1+(a^3-\beta)^2\\3a^2&=3 (-\alpha +a^3+a)^2 \left(3a^2+1\right)-2 (a^3-\beta )3a^2\\6a&=-18 a^4+6 \left(3 a^2+1\right)^2 (-\alpha +a^3+a)+18 a (-\alpha +a^3+a)^2-12 a (a^3-\beta )\endled$
数值求解的结果是$(a,\alpha,\beta)=\left(\{0.300286,-0.141747,-1.02327\right),\left(-0.274586,0.128611,-0.981863\right)$
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发表于 2020-9-5 18:35
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发表于 2020-9-5 23:44
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