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$解：(2)令x_0=t,设A(x_1,y_1),B(x_2,y_2),F(1,0);利用两点式的变形及对偶式$
$\because \angle BFA的角平分线垂直于x轴,\therefore k_{BF}+k_{AF}=0\Rightarrow \cfrac{y_1}{x_1-1}+\cfrac{y_2}{x_2-1}=0$
$\Rightarrow y_1(x_2-1)+y_2(x_1-1)=0\Rightarrow y_1x_2+y_2x_1={\color{Red}y_1+y_2 } $
$由两点式直线方程：\cfrac{y-y_1}{x-x_1} =\cfrac{y_2-y_1}{x_2-x_1}$
$交叉相乘合并同类项，得 y_1x_2-y_2x_1=y(x_2-x_1)+x(y_1-y_2)$
$(x_1,y_1),(x_2,y_2),(t,0)三点共线,得y_1x_2-y_2x_1={\color{Red}t(y_1-y_2) } $
$构造对偶式：y_1x_2+y_2x_1=\cfrac{(y_1x_2)^2-(y_2x_1)^2}{y_1x_2-y_2x_1}=\cfrac{y_1^2x_2^2-y_2^2x_1^2}{t(y_1-y_2)}$
$(x_1,y_1),(x_2,y_2)在椭圆上,有\cfrac{x_1^2}{4}+\cfrac{y_1^2}{3}=1,\Rightarrow \cfrac{x_1^2}{4}=1-\cfrac{y_1^2}{3}$
$\cfrac{x_2^2}{4}+\cfrac{y_2^2}{3}=1,\Rightarrow \cfrac{x_2^2}{4}=1-\cfrac{y_2^2}{3}$
$y_1x_2+y_2x_1=\cfrac{4y_1^2(1-\cfrac{y_2^2}{3})-4y_2^2(1-\cfrac{y_1^2}{3})}{t(y_1-y_2)}=\cfrac{4y_1^2-4y_2^2}{t(y_1-y_2)}=\cfrac{t}{4}(y_1+y_2)$
$即\cfrac{t}{4}=1,t=4$

$19、已知函数f(x)=x+a\cos ax,a\gt 0$
$(1)当a=\sqrt{2}时，求f(x)在区间(0,\cfrac{\pi}{2})上的最大值$
$解：a=\sqrt{2},f(x)x+\sqrt{2}\cos \sqrt{2}x,{f}' (x)=1-2\sin \sqrt{2}x $
$令{f}' (x)=0,1-2\sin \sqrt{2}x =0,解得x=\cfrac{\sqrt{2}\pi}{12},$
$当x\in (0,\cfrac{\sqrt{2}\pi}{12}),{f}' (x)\gt 0,f(x)\nearrow;x\in (\cfrac{\sqrt{2}\pi}{12},\cfrac{\pi}{2}),{f}' (x)\lt 0;f(x)\searrow$
$f(x)在x=\cfrac{\sqrt{2}\pi}{12}有最大值，f(\cfrac{\sqrt{2}\pi}{12})=\cfrac{\sqrt{2}\pi}{12}+\sqrt{2}\cos (\cfrac{\pi}{6})=\cfrac{\sqrt{2}\pi}{12}+\cfrac{\sqrt{6}}{2}$

$15、\triangle ABC中, 角A,B,C所对的边为a,b,c,A=\cfrac{3\pi}{4}.$
$(1)若c=\sqrt{2},b=2,求\sin C$
解：$a^2=b^2+c^2-2bc\cos A=4+2-4\sqrt{2}\cdot (-\cfrac{\sqrt{2}}{2})=10$
$\cfrac{a}{\sin A}=\cfrac{c}{\sin C}\Rightarrow \sin C=\cfrac{c\sin A}{a}=\cfrac{\sqrt{2}\cfrac{\sqrt{2}}{2}}{\sqrt{10}}=\cfrac{\sqrt{10}}{10}$
$(2)\sin B+\sqrt{3}\cos B=2,b=4,求\triangle ABC面积$
$解：\cfrac{1}{2}\sin B+\cfrac{\sqrt{3}}{2}\cos B=1\Rightarrow \sin(B+\cfrac{\pi}{3})=1\Rightarrow B+\cfrac{\pi}{3}=\cfrac{\pi}{2},B=\cfrac{\pi}{6}$

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