Question:

Step 1: Determine the ratio $\frac{b_n^2}{a_n^2}$ using the eccentricity. The eccentricity $e$ of an ellipse is given by the relation $e^2 = 1 - \frac{b_n^2}{a_n^2}$. Given $e = \frac{\sqrt{5}-1}{2}$, we calculate $e^2$: $$e^2 = \left(\frac{\sqrt{5}-1}{2}\right)^2 = \frac{(\sqrt{5})^2 - 2\sqrt{5} + 1^2}{4} = \frac{5 - 2\sqrt{5} + 1}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$$ Now, substitute $e^2$ into the eccentricity relation: $$\frac{b_n^2}{a_n^2} = 1 - e^2 = 1 - \frac{3 - \sqrt{5}}{2} = \frac{2 - (3 - \sqrt{5})}{2} = \frac{2 - 3 + \sqrt{5}}{2} = \frac{\sqrt{5}-1}{2}$$ Step 2: Use the given constraint to establish a ratio involving $x_1$ and $y_1$. The problem states the condition $b_n^2 x_1 = a_n^2 y_1$. Rearranging this equation to find the ratio $\frac{x_1}{y_1}$: $$\frac{x_1}{y_1} = \frac{a_n^2}{b_n^2}$$ Step 3: Combine the ratios to find the relation between $x_1$ and $y_1$. From Step 1, we have $\frac{b_n^2}{a_n^2} = \frac{\sqrt{5}-1}{2}$. From Step 2, we have $\frac{x_1}{y_1} = \frac{a_n^2}{b_n^2}$. Therefore, we can write: $$\frac{x_1}{y_1} = \frac{1}{\frac{b_n^2}{a_n^2}} = \frac{1}{\frac{\sqrt{5}-1}{2}} = \frac{2}{\sqrt{5}-1}$$ To rationalize the denominator, multiply the numerator and denominator by the conjugate $\sqrt{5}+1$: $$\frac{x_1}{y_1} = \frac{2}{\sqrt{5}-1} \cdot \frac{\sqrt{5}+1}{\sqrt{5}+1} = \frac{2(\sqrt{5}+1)}{(\sqrt{5})^2 - 1^2} = \frac{2(\sqrt{5}+1)}{5-1} = \frac{2(\sqrt{5}+1)}{4} = \frac{\sqrt{5}+1}{2}$$ Finally, express the relation between $x_1$ and $y_1$: $$2x_1 = (\sqrt{5}+1)y_1$$