079. 在平面直角坐标系$xoy$中,已知双曲线C:$2{{x}^{2}}-{{y}^{2}}=1$。
(1)设F是C的左焦点,M是C右支上一点。若|MF|=2,求点M的坐标;
(2)设斜率为k(|k|<$\sqrt{2}$)的直线l交C于P、Q两点.若l与圆x2 +y2 =1相切,求证:OP⊥OQ。
(1)解:∵双曲线C:$\dfrac{{{x}^{2}}}{\dfrac{1}{2}}-{{y}^{2}}=1$,
∴左焦点F$(-\dfrac{\sqrt{6}}{2},0)$,
设M(x,y),则|MF|2 =${{(x+\dfrac{\sqrt{6}}{2})}^{2}}+{{y}^{2}}={{(\sqrt{3}x+\dfrac{\sqrt{2}}{2})}^{2}}$,
由M点是右支上一点,知x≥$\dfrac{\sqrt{2}}{2}$,
∴$\left| MF \right|=\sqrt{3}x+\dfrac{\sqrt{2}}{2}=2\sqrt{2}$,得$x=\dfrac{\sqrt{6}}{2}$,
∴,∴M$(\dfrac{\sqrt{6}}{2},\pm \sqrt{2})$。
(2)证明:设直线PQ的方程是y=kx+b,
∵直线PQ与已知圆相切,
∴$\dfrac{\left| b \right|}{\sqrt{{{k}^{2}}+1}}=1$,即b2 =k2 +1。
设P(x1 ,y1 )、Q(x2 ,y2 ),
由$\left\{ \begin{align} & y=kx+b \\ & 2{{x}^{2}}-{{y}^{2}}=1 \\ \end{align} \right.$得(2-k2 )x2 -2kbx-b2 -1=0。
∴$\left\{ \begin{align} & {{x}_{1}}+{{x}_{2}}=\dfrac{2kb}{2-{{k}^{2}}} \\ & {{x}_{1}}\centerdot {{x}_{2}}=\dfrac{-1-{{b}^{2}}}{2-{{k}^{2}}} \\ \end{align} \right.$
∵${{y}_{1}}{{y}_{2}}=(k{{x}_{1}}+b)(k{{x}_{2}}+b)={{k}^{2}}{{x}_{1}}{{x}_{2}}+kb({{x}_{1}}+{{x}_{2}})+{{b}^{2}}$
∴$\overrightarrow{OP}\centerdot \overrightarrow{OQ}={{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}=(1\text{+}{{k}^{2}}{{x}_{1}}{{x}_{2}}\text{+}kb({{x}_{1}}+{{x}_{2}})+{{b}^{2}}$
$\text{=}\dfrac{(1\text{+}{{k}^{2}}\left( -1-{{b}^{2}} \right)}{2-{{k}^{2}}}+\dfrac{2{{k}^{2}}{{b}^{2}}}{2-{{k}^{2}}}+{{b}^{2}}$
$=\dfrac{{{b}^{2}}-{{k}^{2}}-1}{2-{{k}^{2}}}$
∵${{b}^{2}}={{k}^{2}}+1$,∴$\dfrac{{{b}^{2}}-{{k}^{2}}-1}{2-{{k}^{2}}}=0$
∴$\overrightarrow{OP}\centerdot \overrightarrow{OQ}=0$,∴OP⊥OQ。