圆锥曲线15

思路简单，计算量过大的一题，强行堆砌计算量

已知抛物线\(C:y^2=2x\)的焦点为\(F\)，其准线\(l\)与\(x\)轴交于点\(P\)，过点\(P\)的直线与\(C\)交于点\(A,B\)(\(A\)在\(B\)的左侧)

(1)若点\(A\)是线段\(PB\)的中点，求\(A\)的坐标

(2)若直线\(AF\)与\(C\)交于点\(D\)，记\(\triangle BDP\)内切圆的半径为\(r\)，求\(r\)的取值范围.

解

(1)\(P\left(-\dfrac{1}{2},0\right),A\left(\dfrac{y_1^2}{2},y_1\right),B\left(\dfrac{y_2^2}{2},y_2\right)\)

即\(\begin{cases} \dfrac{y_2+0}{2}=y_1\\ \dfrac{y_2^2-1}{4}=\dfrac{y_1^2}{2} \end{cases}\)得\(y_1=\pm\dfrac{\sqrt{2}}{2}\)，则\(A\left(\dfrac{1}{4},\pm\dfrac{\sqrt{2}}{2}\right)\)

(2)\(S_{BDP}=\dfrac{1}{2}(BP+PD+BD)r\)

设\(AP:my-\dfrac{1}{2}=x\),\(A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)\)

联立\(\begin{cases} my-\dfrac{1}{2}=x\\ y^2=2x \end{cases}\)有\(y^2-2my+1=0\)，即$ y^2-2my+1=0$

即\(y_1y_2=1,y_1+y_2=2m,\Delta=4m^2-1>0\)，得\(m\in\left(-\dfrac{1}{2},\dfrac{1}{2}\right)\)，则\(1<x_1<\dfrac{1}{2}<x_2\)

即\(y_2=\dfrac{1}{y_1}\)，即\(y_2^2=\dfrac{1}{y_1^2}\)，即\(2x_2=\dfrac{1}{2x_1}\)，即\(x_2=\dfrac{1}{4x_1}\)

联立，\(AP\)与\(C\):\(\begin{cases} y=\dfrac{y_1-0}{x_1-\dfrac{1}{2}}\left(x-\dfrac{1}{2}\right)\\ y^2=2x \end{cases}\)

整理有:

\[\dfrac{y_1^2}{\left(x_1-\dfrac{1}{2}\right)^2}\left(x-\dfrac{1}{2}\right)^2=2x \]

整理有:

\[2x_1\left(x-\dfrac{1}{2}\right)^2=2x\left(x_1-\dfrac{1}{2}\right) \]

整理有:

\[2x_1\left(x^2-x+\dfrac{1}{4}\right)=2x\left(x_1^2-x_1+\dfrac{1}{4}\right) \]

整理有:

\[2x_1x^2-2xx_1+\dfrac{x_1}{2}=2xx_1^2-2xx_1+\dfrac{x}{2} \]

整理有:

\[2x_1x^2+x\left(2x_1^2+\dfrac{1}{2}\right)+\dfrac{x_1}{2} \]

则\(x_1x_3=\dfrac{1}{4}\)，即\(x_3=\dfrac{1}{4x_1}\)

从而\(x_2=x_3\)，即\(BD\perp\)垂直\(x\)轴

则\(\triangle BPD\)的周长\(2|BP|+2|y_2|\)

则\(\dfrac{1}{2}\left(2|BP|+2|y_2|\right)r=\dfrac{1}{2}\left(x_2+\dfrac{1}{2}\right)\cdot2|y_2|\)

整理有$$r=\dfrac{\left(x_2+\dfrac{1}{2}\right)\cdot|y_2|}{|BP|+|y_2|}$$

即

\(r=\dfrac{\left(x_2+\dfrac{1}{2}\right)\cdot|y_2|}{|y_2|+\sqrt{\left(x_2+\dfrac{1}{2}\right)^2+y_2^2}}\)

\(=\dfrac{1}{\dfrac{1}{x_2+\dfrac{1}{2}}+\sqrt{\dfrac{1}{y_2^2}+\left(x_2+\dfrac{1}{2}\right)^2}}\)

\(=\dfrac{1}{\dfrac{1}{x_2+\dfrac{1}{2}}+\sqrt{\dfrac{1}{2x_2}+\left(x_2+\dfrac{1}{2}\right)^2}}\)

即\(x_2+\dfrac{1}{2}=t\)，则
\(r=\dfrac{1}{\dfrac{1}{t}+\sqrt{\dfrac{1}{2t-1}+\dfrac{1}{t^2}}},t>1\)

不难说明\(\dfrac{1}{\dfrac{1}{t}+\sqrt{\dfrac{1}{2t-1}+\dfrac{1}{t^2}}}\)单调递增

则\(r>\dfrac{1}{1+\sqrt{1+1}}=\sqrt{2}-1\)