向量转换
对于椭圆\(\dfrac{y^2}{a^2}+\dfrac{x^2}{b^2}=1(a>b>0)\)，我们称双曲线\(\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1\)为其伴随双曲线.已知椭圆\(C:\dfrac{y^2}{3}+\dfrac{x^2}{b^2}=1(0<b<\sqrt{3})\)，它的离心率是其伴随双曲线\(\Gamma\)离心率的\(\dfrac{\sqrt{2}}{2}\)倍.
\((1)\) 求椭圆伴随双曲线\(\Gamma\)的方程
\((2)\) 如图，点\(E,F\)分别是\(\Gamma\)的下顶点跟上焦点，过\(F\)的直线\(l\)与\(\Gamma\)上支交与\(A,B\)两点，设\(\triangle ABO\)的面积为\(S,\angle AOB=\theta\).若\(\triangle ABE\)的面积为\(6+3\sqrt{3}\)，求\(\dfrac{S}{\tan\theta}\).
解
\((1)\) \(\dfrac{y^2}{3}-x^2=1\)
\((2)\) 因\(S=\dfrac{1}{2}|AO|\cdot |BO|\sin\theta\)
则\(\dfrac{S}{\tan\theta}=\dfrac{1}{2}|AO|\cdot |BO|\cos\theta=\dfrac{1}{2}\cdot\overrightarrow{OA}\cdot\overrightarrow{OB}\)
设\(A:(x_1,y_1),B:(x_2,y_2),S_{\triangle ABE}=\dfrac{1}{2}|EF|\cdot|x_1-x_2|=\dfrac{\sqrt{3}+2}{2}\cdot|x_1-x_2|\)
又因\(\triangle ABE\)的面积为\(6+3\sqrt{3}\)，所以\(|x_1-x_2|=6\)
设\(AB:y=kx+2\)与\(\Gamma\)联立有
\(x^2(k^2-3)+4kx+1\)
从而\(x_1+x_2=\dfrac{-4k}{k^2-3},x_1x_2=\dfrac{1}{k^2-3}\)
则\(|x_1-x_2|=\sqrt{(x_1+x_2)^2-4x_1x_2}=\sqrt{\dfrac{16k^2-4(k^2-3)}{(k^2-3)^2}}=\sqrt{\dfrac{12k^2+12}{(k^2-3)^2}}\)
从而\((x_1-x_2)^2=\dfrac{12k^2+12}{(k^2-3)^2}=36\)
即\(3k^4-17k^2+26=0\)解得\(k^2=2\)或\(k^2=\dfrac{13}{3}\)
又因\(\Delta>0\)从而舍去\(k^2=\dfrac{13}{3}\)
\(\overrightarrow{OA}\cdot\overrightarrow{OB}=x_1x_2+y_1y_2=x_1x_2(1+k^2)+2k(x_1+x_2)+4=\dfrac{1+k^2-8k^2+4k^2-12}{k^2-3}=\dfrac{-3k^2-11}{k^2-3}=17\)
从而\(\dfrac{S}{\tan\theta}=\dfrac{17}{2}\)