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$$
\dfrac{x2}{2}+y2=1,x\neq\pm \sqrt{2}.$$
\((2)\) (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
\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.
\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$$\left(\dfrac{1}{2}\cos2\theta+\sin2\theta\right)\cdot t^2+\dfrac{1}{2}\left(\cos\theta+\sin\theta\right)\cdot t-\dfrac{13}{16}=0.
|PC|\cdot|PD|=|t_3t_4|=\dfrac{13}{8(1+\sin^2\theta)}.$$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\(\dfrac12\).