25~26高二上·四川绵阳南山中学期中·第14题

$\frac{4(5 – 2\sqrt{5})}{5}$

【题目】在平面直角坐标系中，点$A$在圆$C$：$(x – 1)^2 + (y + 3)^2 = 4$上，点$B$在直线$l$：$2x – y + 4 = 0$上，且$OA \perp OB$，则$\frac{|OB|}{|OA|}$的最小值为______．

【答案】$\frac{4(5 – 2\sqrt{5})}{5}$

【解析】如图，作$AN\bot x$轴于$N$，作$BM\bot x$轴于$M$，

易得$\frac{|OB|}{|OA|} = \frac{|OM|}{|AN|} = \left| \frac{x_B}{y_A} \right|$，

设$A\left( 1 + 2\cos\alpha, -3 + 2\sin\alpha \right)$，

则$k_{OA} = \frac{-3 + 2\sin\alpha}{1 + 2\cos\alpha}$，

$k_{OB} = \frac{1 + 2\cos\alpha}{3 – 2\sin\alpha}$，

$y = \frac{1 + 2\cos\alpha}{3 – 2\sin\alpha} \cdot x$，

$2x – \frac{1 + 2\cos\alpha}{3 – 2\sin\alpha} \cdot x + 4 = 0$，

$\left( 2 – \frac{1 + 2\cos\alpha}{3 – 2\sin\alpha} \right)x = -4$，

$x_B = \frac{-4}{2 – \frac{1 + 2\cos\alpha}{3 – 2\sin\alpha}} $$= \frac{-4(3 – 2\sin\alpha)}{6 – 4\sin\alpha – 1 – 2\cos\alpha} $$= \frac{-4(3 – 2\sin\alpha)}{5 – 4\sin\alpha – 2\cos\alpha}$，

所以$\left| \frac{x_B}{x_A} \right| = \frac{4}{|5 – 4\sin\alpha – 2\cos\alpha|}$$= \frac{4}{|5 – 2\sqrt{5}\sin(\alpha + \varphi)|}$$= \frac{4}{5 + 2\sqrt{5}} = \frac{4(5 – 2\sqrt{5})}{25 – 20} = \frac{4(5 – 2\sqrt{5})}{5}$.