MOOC第六单元作业
1
当\(Y\)点满足过\(CYD\)三点的圆与垂线\(XE\)相切时，\(\angle CYD\)最大
2
(1)
令\(\angle ADC=\varphi,\angle DAB=\theta,b=\sqrt{2}a\)。则有\(B(-a+b\cos\theta,-b\sin\theta),C(a-b\cos\varphi,b\sin\varphi).\)
有：
\[\begin{aligned}4a^{2}&=BC^2\\
&=(2a-b(\cos\theta+\cos\varphi))^2+b^2(\sin\theta+\cos\varphi)^2\\&=(2a-2b\cos\frac{\theta+\varphi}2\cos\frac{\theta-\varphi}2)^2+4b^2\sin^2\frac{\theta+\varphi}2\cos^2\frac{\theta-\varphi}2\\&=4a^2-8ab\cos\frac{\theta+\varphi}2\cos\frac{\theta-\varphi}2+4b^2\cos^2\frac{\theta+\varphi}2\cos^2\frac{\theta-\varphi}2+4b^2\sin^2\frac{\theta+\varphi}2\cos^2\frac{\theta-\varphi}2\\&=4a^2-8ab\cos\frac{\theta+\varphi}2\cos\frac{\theta-\varphi}2+4b^2\cos^2\frac{\theta-\varphi}2,\end{aligned}\]
\[
b\cos\frac{\theta-\varphi}2=2a\cos\frac{\theta+\varphi}2
\]
设\(P\)点坐标为\((x,y)\)，有
\[\left\{\begin{array}{ll}
x=\frac b2(\cos\theta-\cos\varphi) \\
y=-\frac b2(\sin\theta-\sin\varphi)
\end{array}\right.\]
可以得到：
\[\begin{aligned}x^{2}+y^{2}&=\frac{b^{2}}{4}(2-2\cos(\theta-\varphi)) \\
&=\frac{b^{2}}{2}(1-\cos(\theta-\varphi)) \\
&=b^{2}\sin^{2}\frac{\theta-\varphi}{2}\\
x^{2}-y^{2}&=\frac{b^{2}}{4}(\cos2\theta+\cos2\varphi-2\cos(\theta+\varphi)) \\
&=\frac{b^{2}}{4}(2\cos(\theta+\varphi)\cos(\theta-\varphi)-2\cos(\theta+\varphi))\\
&=\frac{b^{2}}{2}\cos(\theta+\varphi)(\cos(\theta-\varphi)-1)\end{aligned}\]
因此
\[
\cos(\theta+\varphi)=\frac{y^2-x^2}{y^2+x^2},\cos\frac{\theta+\varphi}2=\frac y{\sqrt{x^2+y^2}}
\]
\[
\cos\frac{\theta-\varphi}2=\frac{2a}b\frac y{\sqrt{x^2+y^2}}=\sqrt{2}\frac y{\sqrt{x^2+y^2}}
\]
\[
\sin^2\frac{\theta-\varphi}2=1-\frac{2y^2}{x^2+y^2}=\frac{x^2-y^2}{x^2+y^2}
\]
\(P\) 的轨迹方程为：
\[
(x^2+y^2)^2=2a^2(x^2-y^2)
\]
又有
\[
x=b\sin\frac{\theta+\varphi}2\sin\frac{\varphi-\theta}2,y=b\cos\frac{\theta+\varphi}2\sin\frac{\varphi-\theta}2
\]
令\(t=\frac xy=\tan\frac{\theta+\varphi}2\) ,则\(\cos \frac {\theta - \varphi }2= \sqrt {2}\frac 1{\sqrt {1+ t^2}}\)，从而 \(P\) 的参数方程为
\[\begin{cases}x&=\sqrt{2}a\frac{t\sqrt{t^2-1}}{t^2+1}\\[2ex]y&=\sqrt{2}a\frac{\sqrt{t^2-1}}{t^2+1}\end{cases}\]
(2)
将\(P\)的轨迹方程求导得\(2( x^2+ y^2) ( 2x+ 2y\cdot y^{\prime }) = 2a^2( 2x- 2y\cdot y^{\prime })\)，即\(y^\prime=\frac xy\frac{a^2-x^2-y^2}{a^2+x^2+y^2}\)。故过\(P(x_0,y_0)\)的法线方程为\(y-y_0=\frac{y_0}{x_0}\frac{x_0^2+y_0^2+a^2}{x_0^2+y_0^2-a^2}(x-x_0)\) ,法线与\(x\)轴交点\(Q\)的坐标为\(\left(\frac{2a^2x_0}{a^2+x_0^2+y_0^2},0\right)\)
记\(\angle POQ=\alpha,\angle OPQ=\beta\)，则有\(\sin\alpha=\frac{y_{0}}{\sqrt{x_{0}^{2}+y_{0}^{2}}}\)。由正弦定理\(\frac {| PQ| }{\sin \alpha }= \frac {| OQ| }{\sin \beta }\)，有
\[\begin{aligned}\sin^{2}\beta&=\frac{|OQ|^{2}}{|PQ|^{2}}\sin^{2}\alpha \\
&=\frac{\frac{4a^{4}x_{0}^{2}}{(a^{2}+x_{0}^{2}+y_{0}^{2})^{2}}\frac{y_{0}^{2}}{x_{0}^{2}+y_{0}^{2}}}{(\frac{2a^{2}x_{0}}{a^{2}+x_{0}^{2}+y_{0}^{2}}-x_{0})^{2}+y_{0}^{2}} \\
&=\frac{4a^{4}x_{0}^{2}\frac{y_{0}^{2}}{x_{0}^{2}+y_{0}^{2}}}{(a^{2}-x_{0}^{2}-y_{0}^{2})^{2}x_{0}^{2}+(a^{2}+x_{0}^{2}+y_{0}^{2})^{2}y_{0}^{2}} \\
&=\frac{4a^{4}x_{0}^{2}y_{0}^{2}}{(x_{0}^{2}+y_{0}^{2})(a^{4}+2a^{2}(y_{0}^{2}-x_{0}^{2})+(x_{0}^{2}+y_{0}^{2})^{2})} \\
&=\frac{4x_{0}^{2}y_{0}^{2}}{(x_{0}^{2}+y_{0}^{2})^{2}}\\
&=4\sin^{2}\alpha\cos^{2}\alpha \end{aligned}\]
故\(\beta=2\alpha,\angle PQD=\angle POQ+\angle OPQ=3\alpha\)