圆锥曲线14

数据不好的斜率定值找定点、隐圆

已知椭圆\(C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)，直线\(l:y=\dfrac{1}{2}x+t\)过\(C\)的左定点与上定点，且\(l\)与两坐标轴围成的三角形面积为\(1\)

(1)求\(C\)的标准方程

(2)已知点\(P\left(-\sqrt{2},-\dfrac{\sqrt{2}}{2}\right),A,B\)(异于\(P\)点)是椭圆\(C\)上不同的两点，且\(PA\perp PB\)，过点\(P\)做\(AB\)的垂线，垂足为\(M\)，求\(M\)到直线\(l\)距离的最大值.

解

(1)\(\dfrac{x^2}{4}+y^2=1\)

(2)当直线\(AB\)的斜率存在时，设\(AB\)方程\(y=kx+m,A(x_1,y_1),B(x_2,y_2)\)

联立\(\begin{cases} \dfrac{x^2}{4}+y^2=1\\ y=kx+m \end{cases}\)有\((1+4k^2)+8km+4m^2-4=0\)

即\(x_1+x_2=-\dfrac{8km}{1+4k^2},x_1x_2=\dfrac{4m^2-4}{1+4k^2}\)

\(y_1+y_2=k(x_1+x_2)+2m,y_1y_2=kx_1x_2+km(x_1+x_2)+m^2\)

由\(PA\perp PB\)得\(\overrightarrow{PA}\cdot\overrightarrow{PB}=0\)，

有

\(\left(x_1+\sqrt{2},y_1+\dfrac{1}{\sqrt{2}}\right)\cdot\left(x_2+\sqrt{2},y_2+\dfrac{1}{\sqrt{2}}\right)\)

\(=x_1x_2+\sqrt{2}(x_1+x_2)+2+y_1y_2+\dfrac{1}{\sqrt{2}}(y_1+y_2)+\dfrac{1}{2}\)

\(=x_1x_2+\sqrt{2}(x_1+x_2)+2+kx_1x_2+km(x_1+x_2)+m^2+\dfrac{1}{\sqrt{2}}[k(x_1+x_2)+2m]+\dfrac{1}{2}\)

\(=x_1x_2(1+k)+\left(\sqrt{2}+km+\dfrac{k}{\sqrt{2}}\right)(x_1+x_2)+m^2+\sqrt{2}m+\dfrac{5}{2}\)

代入\(x_1+x_2=-\dfrac{8km}{1+4k^2},x_1x_2=\dfrac{4m^2-4}{1+4k^2}\)有

\[(4m^2-4)(1+k)-8km\left(\sqrt{2}+km+\dfrac{k}{\sqrt{2}}\right)+\left(m^2+\sqrt{2}m+\dfrac{5}{2}\right)(1+4k^2)=0 \]

整理有

\[5m^2+\sqrt{2}m-4k-\dfrac{3}{2}-8\sqrt{2}km-8k^2m^2+10k^2=0 \]

做因式分解有

\[\left(4m-3\sqrt{2}\right)^2=\left(m-\dfrac{3\sqrt{2}}{2}\right)^2 \]

得\(m=\sqrt{2}\left(k-\dfrac{1}{2}\right)\)或\(m=\dfrac{3\sqrt{2}}{5}\left(k+\dfrac{1}{2}\right)\)

因\(AB\)不过点\(P\),则\(m=\dfrac{3\sqrt{2}}{5}\left(k+\dfrac{1}{2}\right)\)

此时\(AB:y=k\left(x+\dfrac{3\sqrt{2}}{5}\right)+\dfrac{3\sqrt{2}}{10}\)，过定点\(Q\left(-\dfrac{3\sqrt{2}}{5},\dfrac{3\sqrt{2}}{10}\right)\)

当\(AB\)斜率不存在时，同样得到定点\(Q\left(-\dfrac{3\sqrt{2}}{5},\dfrac{3\sqrt{2}}{10}\right)\)

由\(PM\perp PB\)，\(AB\)过定点\(Q\)，则\(PQ\)是定值，从而\(M\)点的轨迹是以\(PQ\)中点为圆心，\(\dfrac{PQ}{2}\)为半径的圆

则\(M\)所在圆方程是:\(\left(x+\dfrac{4\sqrt{2}}{5}+x\right)^2+\left(y+\dfrac{\sqrt{2}}{10}\right)^2=\dfrac{2}{5}\)

则\(M\)到直线\(l\)距离的最大值为\(\dfrac{\left|-4\sqrt{2}+\dfrac{\sqrt{2}}{5}+2\right|}{\sqrt{5}}+\dfrac{\sqrt{10}}{5}=\dfrac{10\sqrt{5}+2\sqrt{10}}{25}\)