資料不好的斜率定值找定點、隱圓
已知橢圓\(C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)，直線\(l:y=\dfrac{1}{2}x+t\)過\(C\)的左定點與上定點，且\(l\)與兩座標軸圍成的三角形面積為\(1\)
(1)求\(C\)的標準方程
(2)已知點\(P\left(-\sqrt{2},-\dfrac{\sqrt{2}}{2}\right),A,B\)(異於\(P\)點)是橢圓\(C\)上不同的兩點，且\(PA\perp PB\)，過點\(P\)做\(AB\)的垂線，垂足為\(M\)，求\(M\)到直線\(l\)距離的最大值.
解
(1)\(\dfrac{x^2}{4}+y^2=1\)
(2)當直線\(AB\)的斜率存在時，設\(AB\)方程\(y=kx+m,A(x_1,y_1),B(x_2,y_2)\)
聯立\(\begin{cases} \dfrac{x^2}{4}+y^2=1\\ y=kx+m \end{cases}\)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
整理有
做因式分解有
得\(m=\sqrt{2}\left(k-\dfrac{1}{2}\right)\)或\(m=\dfrac{3\sqrt{2}}{5}\left(k+\dfrac{1}{2}\right)\)
因\(AB\)不過點\(P\),則\(m=\dfrac{3\sqrt{2}}{5}\left(k+\dfrac{1}{2}\right)\)
此時\(AB:y=k\left(x+\dfrac{3\sqrt{2}}{5}\right)+\dfrac{3\sqrt{2}}{10}\)，過定點\(Q\left(-\dfrac{3\sqrt{2}}{5},\dfrac{3\sqrt{2}}{10}\right)\)
當\(AB\)斜率不存在時，同樣得到定點\(Q\left(-\dfrac{3\sqrt{2}}{5},\dfrac{3\sqrt{2}}{10}\right)\)
由\(PM\perp PB\)，\(AB\)過定點\(Q\)，則\(PQ\)是定值，從而\(M\)點的軌跡是以\(PQ\)中點為圓心，\(\dfrac{PQ}{2}\)為半徑的圓
則\(M\)所在圓方程是:\(\left(x+\dfrac{4\sqrt{2}}{5}+x\right)^2+\left(y+\dfrac{\sqrt{2}}{10}\right)^2=\dfrac{2}{5}\)
則\(M\)到直線\(l\)距離的最大值為\(\dfrac{\left|-4\sqrt{2}+\dfrac{\sqrt{2}}{5}+2\right|}{\sqrt{5}}+\dfrac{\sqrt{10}}{5}=\dfrac{10\sqrt{5}+2\sqrt{10}}{25}\)