Question:

Step 1: Rewrite the given line in slope-intercept form. The line $2px + y\sqrt{1-p^2} = 1$ can be rearranged to isolate $y$: $$y = \frac{1 - 2px}{\sqrt{1-p^2}}$$ This gives us the slope-intercept form $y = mx + c$ where: $$m = \frac{-2p}{\sqrt{1-p^2}} \quad \text{and} \quad c = \frac{1}{\sqrt{1-p^2}}$$ Step 2: Apply the condition of tangency to an ellipse. For a line $y = mx + c$ to be tangent to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, the tangency condition is: $$c^2 = a^2m^2 + b^2$$ Step 3: Calculate $c^2$ and $m^2$. From Step 1, we compute: $$c^2 = \frac{1}{1-p^2}$$ $$m^2 = \frac{4p^2}{1-p^2}$$ Step 4: Substitute into the tangency condition. Substituting $c^2$ and $m^2$ into the tangency condition: $$\frac{1}{1-p^2} = a^2 \cdot \frac{4p^2}{1-p^2} + b^2$$ Multiplying both sides by $(1-p^2)$: $$1 = 4a^2p^2 + b^2(1-p^2)$$ Step 5: Expand and rearrange the tangency equation. Expanding the right side: $$1 = 4a^2p^2 + b^2 - b^2p^2$$ Rearranging: $$1 = b^2 + p^2(4a^2 - b^2)$$ Step 6: Apply the condition that the equation holds for all $p \in (-1,1) - \{0\}$. For this equation to be true for all values of $p$ in the given domain, the coefficient of $p^2$ must be zero, and the constant term must equal 1: $$b^2 = 1 \quad \text{and} \quad 4a^2 - b^2 = 0$$ From the second equation: $$4a^2 = b^2 = 1 \implies a^2 = \frac{1}{4}$$ Step 7: Identify the major and minor axes. Since $a^2 = \dfrac{1}{4} < b^2 = 1$, we have $a < b$. This means the major axis is along the $y$-axis, and $b$ is the semi-major axis. Step 8: Calculate the eccentricity. For an ellipse with the major axis along the $y$-axis, the eccentricity is: $$e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1/4}{1}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$ **Final Answer:** The eccentricity of the ellipse is $\boxed{\dfrac{\sqrt{3}}{2}}$, which corresponds to **Option 4**.