在平面直角坐标系\(xOy\)中,动点\(M\)与两定点\((-\sqrt2,0)\),\((\sqrt2,0)\)连线的斜率之积为\(-\dfrac 12\).
\((1)\) 求动点\(M\)的轨迹\(E\)的方程;
\((2)\) 过点\((2,2)\)作\(E\)的两条切线,切点分别为\(A,B\),过点\(P\left(\dfrac{1}{2},\dfrac14\right)\)作直线与\(E\)交于\(C,D(\)异于\(A,B)\)两点,且满足\(|PA|\cdot |PB|=|PC|\cdot |PD|\).
(I) 证明: \(P,A,B\)三点共线;
(II) 求直线\(CD\)的斜率.
解析:
\((1)\) 设动点\(M\left(x,y\right)\),其中\(x\neq \pm\sqrt{2}\),则\[ \dfrac{y-0}{x-\sqrt{2}}\cdot \dfrac{y-0}{x-\left(-\sqrt{2}\right)}=-\dfrac{1}{2}.\]整理可得\(M\)的轨迹\(E\)的方程为\[ \dfrac{x^2}{2}+y^2=1,x\neq\pm \sqrt{2}.\]
\((2)\) (I) \(\qquad\)法一 \(\qquad\)点\((2,2)\)关于曲线\(E\)的切点弦方程为\[ l_{AB}:x+2y=1.\]显然\(P\)点的坐标满足该直线方程,因此\(P,A,B\)三点共线得证.
法二 \(\qquad\) 设点\(A,B\)的坐标分别为\((x_1,y_1)\),\((x_2,y_2)\),记\(Q(2,2)\),则直线\(QA\)的方程为
\[l_{QA}: \dfrac{x_1x}{2}+y_1y=1.\]证明如下,考虑方程组\[(\ast)\qquad \dfrac{x_1x}{2}+y_1y=1, \dfrac{x^2}{2}+y^2=1,\dfrac{x_1^2}{2}+y_1^2=1. \] 联立以上三式可得\[ \dfrac{(x-x_1)^2}{2}+\left(y-y_1\right)^2=0.\]因此不等式组\((\ast)\)有且仅有一解\((x,y)=(x_1,y_1)\),所以直线\(l_{QA}\)与曲线\(E\)相切于点\(A\),同理可证\[l_{QB}: \dfrac{x_2x}{2}+y_2y=1.\]由于\(Q\)同时满足直线\(l_{QA}\)与\(l_{QB}\)的方程,即有\[ x_1+2y_1=1,x_2+2y_2=1. \]
所以\(A(x_1,y_1)\),\(B(x_2,y_2)\)均满足直线方程\(x+2y=1\),因此直线\(AB\)的方程为\[l_{AB}: x+2y=1.\]显然\(P\left(\dfrac{1}{2},\dfrac{1}{4}\right)\)满足该直线方程,因此\(P,A,B\)三点共线得证.
(II) \(\qquad\) 根据\(AB\)直线方程,易得直线\(AB\)的一个参数方程如下\[ \begin{cases} x=\dfrac{1}{2}-\dfrac{2\sqrt{5}}{5}t, y=\dfrac{1}{4}+\dfrac{\sqrt{5}}{5}t, \end{cases} \text{$t$为参数}.\]
将该参数方程代入曲线\(E\)的直角坐标方程并整理可得\[\dfrac{3}{5}t^2-\dfrac{\sqrt{5}}{10}t-\dfrac{13}{16}=0.\]
设\(A,B\)两点对应的参数分别为\(t_1,t_2\),则\(t_1,t_2\)是上述方程的解,则\[|PA|\cdot |PB|=|t_1t_2|=\dfrac{65}{48}.\]
设直线\(CD\)的倾斜角为\(\theta\),则直线\(CD\)的一个参数方程为\[ \begin{cases} x=\dfrac{1}{2}+t\cos\theta,\\ y=\dfrac{1}{4}+t\sin\theta, \end{cases}\text{$t$为参数}.\]
将该参数方程代入曲线\(E\)的直角坐标方程并整理可得\[\left(\dfrac{1}{2}\cos^2\theta+\sin^2\theta\right)\cdot t^2+\dfrac{1}{2}\left(\cos\theta+\sin\theta\right)\cdot t-\dfrac{13}{16}=0. \]设\(C,D\)两点对应的参数分别为\(t_3,t_4\),则\(t_3,t_4\)是上述方程的解,则\[ |PC|\cdot|PD|=|t_3t_4|=\dfrac{13}{8(1+\sin^2\theta)}.\]又因为\(|PA|\cdot |PB|=|PC|\cdot |PD|\),所以\[\dfrac{65}{48}= \dfrac{13}{8(1+\sin^2\theta)}.\]解得\(\sin\theta=\dfrac{\sqrt{5}}{5}\),又因为\(CD\)直线与\(AB\)直线不重合,因此直线\(CD\)的斜率为\(\tan\theta=\dfrac{1}{2}\).
由于\[|PA|\cdot |PB|=|PC|\cdot |PD|.\]所以\(A,B,C,D\)四点共圆,因此直线\(AB,CD\)斜率互为相反数,因此\(CD\)的斜率为\(\dfrac12\).