Question

Sol. [Key Concept: Take the square's side as the unknown; the curve $y^{2} = kx$ through the corners yields a quadratic whose side-ratio is the golden ratio $\frac{1 + \sqrt{5}}{2}$.]

$$y^{2} = k^{2}x \Rightarrow y^{2} = 4\left( \frac{k^{2}}{4} \right)x$$

$$B \equiv \left( \frac{k^{2}}{4}t_{1}^{2},\frac{k^{2}}{2}t_{1} \right),F\left( \frac{k^{2}}{4}t_{2}^{2},\frac{k^{2}}{2}t_{2} \right)$$

$$\begin{matrix} \frac{k^{2}}{4}t_{1}^{2} & \ = \frac{k^{2}}{2}t_{1} \Rightarrow t_{1} = 2 \\ \frac{k^{2}}{2}t_{2} & \ = \frac{k^{2}}{4}\left( t_{2}^{2} - t_{1}^{2} \right) \\ 2t_{2} & \ = t_{2}^{2} - 4 \Rightarrow t_{2}^{2} - 2t_{2} - 4 = 0 \\ t_{2} & \ = 1 + \sqrt{5} \\ \frac{FG}{BC} & \ = \frac{t_{2}}{t_{1}} = \frac{1 + \sqrt{5}}{2} \end{matrix}$$