Ans: (B)
Hint \(: \because\left(\frac{a}{r}\right)^{2}+a^{2}=a^{2} r^{2} \quad(r>1), a \neq 0 \Rightarrow r^{4}-r^{2}-1=0 \Rightarrow r^{2}=\frac{1 \pm \sqrt{5}}{2} \Rightarrow r=\pm \sqrt{\frac{\sqrt{5}+1}{2}}\)
$$
\Rightarrow r=\sqrt{\frac{\sqrt{5}+1}{2}}(r>1), \frac{1}{r}=\sqrt{\frac{\sqrt{5}-1}{2}}\left(\because \alpha+\beta=90^{\circ} \Rightarrow \tan \alpha=\cot \beta=\frac{1}{\tan \beta}\right)
$$
