切点弦与阿基米德三角形
已知\(F\)是抛物线\(C:x^2=4y\)与椭圆\(\dfrac{y^2}{a^2}+\dfrac{x^2}{b^2}=1(a>b>1)\)的的公共焦点，椭圆上的点\(M\)到点\(F\)的距离的最大值为\(3\)
\((1)\) 求椭圆的方程
\((2)\) 过点\(M\)作\(C\)的两条切线，记切点分别为\(A,B\)求\(\triangle MAB\)面积的最大值
\((1)\) \(\dfrac{y^2}{4}+\dfrac{x^2}{3}=1\)
\((2)\) 设\(A(x_1,y_1),B(x_2,y_2),M(x_0,y_0)\)则\(y=\dfrac{x^2}{4},y^{\prime}(x)=\dfrac{x}{2}\)
则\(AM:y-y_1=\dfrac{x_1}{2}(x-x_1)\)
即\(2y-2y_1=x_1x-x_1^2\)
即\(2y-2y_1=x_1x-4y_1\)
即\(2y+2y_1-x_1x=0\)
同理\(BM:2y+2y_2-x_2x=0\)
而\(M(x_0,y_0)\)在\(BM,AM\)上
从而\(\begin{cases}2y_0+2y_1-x_1x_0=0\\2y_0+2y_2-x_2x_0=0\end{cases}\)
从而\(AB\)直线\(2y_0+2y-x_0x=0\)
点\(M\)到\(AB\)的距离\(d=\dfrac{|2y_0+2y_0-x_0^2|}{\sqrt{4+x_0^2}}=\dfrac{|4y_0-x_0^2|}{\sqrt{x_0^2+4}}\)
\(AB\)与抛物线联立有:\(x^2=4\cdot \dfrac{1}{2}(x_0x-2y_0)\)
即\(x^2-2x_0x+4y_0=0\)
则\(x_1+x_2=2x_0,x_1x_2=4y_0\)
从而\(|AB|=\sqrt{1+\left(\dfrac{x_0}{2}\right)^2}\sqrt{(x_1+x_2)^2-4x_1x_2}=\sqrt{1+\dfrac{x_0^2}{4}}\sqrt{4x_0^2-16y_0}=\sqrt{(x_0^2+4)(x_0^2-4y_0)}\)
则\(S=\dfrac{1}{2}|AB|\cdot d=\dfrac{1}{2}|x_0^2-4y_0|\sqrt{x_0^2-4y_0}=\dfrac{1}{2}|x_0^2-4y_0|^{\frac{3}{2}}\)
又因\(M\)在椭圆上，所以\(x_0^2=3-\dfrac{3y_0^2}{4}\)
所以\(x_0^2-4y_0=3-\dfrac{3y_0^2}{4}-4y_0,y_0\in[-2,2]\)
对称轴\(y_0=-\dfrac{8}{3}\)不难得到\(y=-2\)取\(max=8\)，则\(S_{\max}=2\sqrt2\)