每日一题_191206

已知抛物线(y^2=4x)的焦点为(F),( riangle ABC)的三个顶点都在抛物线上,且(overrightarrow{FB}+overrightarrow{FC}=overrightarrow{FA}).
((1)) 证明:直线(BC)恒过一定点;
((2)) 判断( riangle ABC)可否是锐角三角形,并说明理由.
解析:
((1)) 易知(F(1,0)),设$$
Bleft(4t_1^{2,4t_1 ight),Cleft(4t_2}2,4t_2 ight).$$
由于(overrightarrow{FB}+overrightarrow{FC}=overrightarrow{FA}),若设(A(x_A,y_A)),则$$
left(4t_1^2+4t_22-2,4t_1+4t_2 ight)=left(x_A-1,y_A ight).$$从而$$
egin{cases}
& x_A=4t_1^2+4t_22-1,
& y_A=4t_1+4t_2,
end{cases}

[由于$y_A^2=4x_A$,所以]

left(4t_1+4t_2
ight)^2=4left(4t_1^2+4t_2^2-1
ight).$$解得$t_1t_2=-dfrac{1}{8}$.从而$B,C$点横坐标乘积为定值,即$$4t_1^2cdot 4t_2^2=dfrac{1}{4},$$结合抛物线的几何平均性质可知$BC$直线恒过定点$left(dfrac{1}{2},0
ight)$.证毕
$(2)$ 显然$	riangle BCFcong 	riangle  BCA$,因此只需研究$	riangle BCF$,记$B(x_B,y_B),C(x_C,y_C)$,结合$(1)$有$$
x_Bx_C=dfrac{1}{4},y_By_C=-2.$$此时$$
egin{cases}
& |BC|^2=left(x_B-x_C
ight)^2+left(y_B-y_C
ight)^2,\
& |BF|^2=left(x_B+1
ight)^2,\
& |CF|^2=left(x_C+1
ight)^2,
end{cases}
$$
由于$$
|BC|^2-|BF|^2-|CF|^2=2(x_B+x_C)+dfrac{3}{2}>0,$$因此$	riangle BCF$恒为钝角三角形.从而$	riangle ABC$不可能为锐角三角形.