圆锥曲线17

双切线，同构处理，面积范围，特别地表示直线

已知抛物线\(C_1:y^2=2px(p>0)\)的准线与半椭圆\(C_2:\dfrac{x^2}{4}+y^2=1(x\leq 0)\)相交于\(A,B\)两点，且\(|AB|=\sqrt{3}\)

(1)求抛物线方程

(2)若点\(P\)是半椭圆\(C_2\)上一动点，过点\(P\)作抛物线\(C_1\)两条切线，切点分别为\(CD\)，求\(\triangle PCD\)面积的取值范围

解

(1)\(y^2=4x\)

(2)设\(C\left(\dfrac{y_1^2}{y_1}\right),D\left(\dfrac{y_2^2}{4},y_2\right),P(x_0,y_0),y^{\prime}=\dfrac{1}{\sqrt{x}}\)

\(k_{PC}=\dfrac{y_1-y_0}{\dfrac{y_2^2}{4}-x_0}=\dfrac{1}{\sqrt{\dfrac{y_1^2}{4}}}=\dfrac{2}{y_1}\)

即\(y_1^2-y_0y_1=\dfrac{y_1^2}{2}-2x_0\)

即\(y_1^2-2y_0y_1+4x_0=0\)

同理,\((x_2,y_2)\)也适合上述方程

则\(y_1,y_2\)是方程\(y^2-2y_0y+4x_0=0\)的两个根

则\(y_1+y_2=2y_0,y_1y_2=4x_0\)

\(k_{CD}=\left(\dfrac{y_2-y_1}{\dfrac{(y_2-y_1)(y_2+y_1)}{4}}\right)=\dfrac{4}{y_2+y_1}=\dfrac{2}{y_0}\)

取\(CD\)的中点\(E\left(\dfrac{y_1^2+y_2^2}{8},\dfrac{y_1+y_2}{2}\right)=\left(\dfrac{\left[(y_1+y_2)^2-2y_1y_2\right]}{8},y_0\right)=\left(\dfrac{y_0^2}{2}-x_0,y_0\right)\)

则\(CD:y-y_0=\dfrac{2}{y_0}\left(x+x_0-\dfrac{y_0^2}{2}\right)\)

整理有\(2x-yy_0+2x_0=0\)

点\(P\)到\(CD\)的距离\(d=\dfrac{\left|2x_0-y_0^2+2x_0\right|}{\sqrt{y_0^2+4}}\)

\(CD=\sqrt{\left(\dfrac{y_1^2-y_2^2}{4}\right)^2+(y_1-y_2)^2}\)

\(=\sqrt{\dfrac{1}{16}(y_1-y_2)^2(y_1+y_2)^2+(y_1-y_2)^2}\)

\(=\sqrt{(y_1-y_2)^2\left[\dfrac{1}{16}(y_1+y_2)^2+1\right]}\)

\(=\sqrt{\left[(y_1+y_2)^2-4y_1y_2\right]\left(\dfrac{y_0^2}{4}+1\right)}\)

\(=\sqrt{(y_0^2-4x_0)(y_0^2+4)}\)

\(S=\dfrac{1}{2}|CD|d=\dfrac{1}{2}\sqrt{(y_0^2-4x_0)(y_0^2+4)}\cdot\dfrac{\left|2x_0-y_0^2+2x_0\right|}{\sqrt{y_0^2+4}}\)

\(=\dfrac{|2x_0-y_0^2+2x_0|\sqrt{y_0^2-4x_0}}{2}\)

\(=\dfrac{|4x_0-y_0^2|\sqrt{y_0^2-4x-0}}{2}\)

\(=\dfrac{\left(y_0^2-4x_0\right)^{\frac{3}{2}}}{2}\)

\(=\dfrac{\left(1-\dfrac{x_0^2}{4}-4x_0\right){\frac{3}{2}}}{2}\)

因\(x_0\in(-2,0)\)，则\(S\in\left(\dfrac{1}{2},8\sqrt{2}\right)\).