每日一题[871]圆中的弦

如图,已知动直线$l$经过点$P\left( {4,0} \right)$,交抛物线${y^2} = 2ax$($a > 0$)于$A$、$B$两点.坐标原点$O$是$PQ$的中点,设直线$AQ$、$BQ$的斜率分别为${k_{AQ}}$、${k_{BQ}}$.

(1)证明:${k_{AQ}} + {k_{BQ}} = 0$;

(2)当$a = 2$时,是否存在垂直于$x$轴的直线$l'$,被以$AP$为直径的圆截得的弦长为定值?若存在,请求出直线$l'$的方程;若不存在,请说明理由.



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分析与解 (1)$P\left( {4,0} \right)$、$Q\left( { - 4 , 0} \right)$.设$A\left( {{x_1},{y_1}} \right)$,$B\left( {{x_2} ,{y_2}} \right)$,直线$AB$为$x = my + 4$,则
$${k_{AQ}} + {k_{BQ}} = 0,$$即 $$ \dfrac{{{y_1}}}{{{x_1} + 4}} + \dfrac{{{y_2}}}{{{x_2} + 4}} = 0,$$ 也即$${y_1}\left( {{x_2} + 4} \right) + {y_2}\left( {{x_1} + 4} \right) = 0$$等价于$$ {y_1}\left( {m{y_2} + 8} \right) + {y_2}\left( {m{y_1} + 8} \right) = 0,$$于是只需证$$ m{y_1}{y_2} + 4\left( {{y_1} + {y_2}} \right) = 0.$$
联立直线与抛物线有$${y^2} - 2amy - 8a = 0,$$所以$${y_1}{y_2} = - 8a,\; {y_1} + {y_2} = 2am.$$因此原命题得证.

(2)假设存在这样的直线$l'$:$x = t$,则$AP$的中点$M$到$l'$的距离为$\dfrac{{{x_1} + 4}}{2} - t$.而$$\left| {AP} \right| = \sqrt {{{\left( {{x_1} - 4} \right)}^2} + {y_1}^2} ,$$所以半径为$$\dfrac{1}{2}\left| {AP} \right| = \dfrac{1}{2}\sqrt {{{\left( {{x_1} - 4} \right)}^2} + 4{x_1}} ,$$所以弦长为$$l = \sqrt {{{\left( {{x_1} - 4} \right)}^2} + 4{x_1} - {{\left( {{x_1} + 4 - 2t} \right)}^2}} = 2\sqrt {\left( {t - 3} \right){x_1} + 4t - {t^2}} $$为定值.

所以当$t = 3$时符合题意,此时直线$l'$的方程为$x = 3$.

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