已知抛物线\(y^2=4x\)的焦点为\(F\),\(\triangle ABC\)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\(\triangle ABC\)可否是锐角三角形,并说明理由.
解析:
\((1)\) 易知\(F(1,0)\),设\[ B\left(4t_1^2,4t_1\right),C\left(4t_2^2,4t_2\right).\]
由于\(\overrightarrow{FB}+\overrightarrow{FC}=\overrightarrow{FA}\),若设\(A(x_A,y_A)\),则\[ \left(4t_1^2+4t_2^2-2,4t_1+4t_2\right)=\left(x_A-1,y_A\right).\]从而\[ \begin{cases} & x_A=4t_1^2+4t_2^2-1,\\ & y_A=4t_1+4t_2, \end{cases} \]
由于\(y_A^2=4x_A\),所以\[ \left(4t_1+4t_2\right)^2=4\left(4t_1^2+4t_2^2-1\right).\]解得\(t_1t_2=-\dfrac{1}{8}\).从而\(B,C\)点横坐标乘积为定值,即\[4t_1^2\cdot 4t_2^2=\dfrac{1}{4},\]结合抛物线的几何平均性质可知\(BC\)直线恒过定点\(\left(\dfrac{1}{2},0\right)\).证毕
\((2)\) 显然\(\triangle BCF\cong \triangle BCA\),因此只需研究\(\triangle BCF\),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\[ \begin{cases} & |BC|^2=\left(x_B-x_C\right)^2+\left(y_B-y_C\right)^2,\\ & |BF|^2=\left(x_B+1\right)^2,\\ & |CF|^2=\left(x_C+1\right)^2, \end{cases} \]
由于\[ |BC|^2-|BF|^2-|CF|^2=2(x_B+x_C)+\dfrac{3}{2}>0,\]因此\(\triangle BCF\)恒为钝角三角形.从而\(\triangle ABC\)不可能为锐角三角形.