1. **Stating the problem:** Determine the lines parallel to the x-axis that cut a segment on the ellipse $$\frac{x^2}{4} + \frac{y^2}{16} = 1$$ whose length equals the semi-focal distance. 2. **Recall the ellipse parameters:** The ellipse is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a^2=4$$ and $$b^2=16$$. 3. **Calculate the semi-focal distance $$c$$:** $$c = \sqrt{b^2 - a^2} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}$$. 4. **Lines parallel to x-axis have equation $$y = k$$.** We want to find $$k$$ such that the chord length cut by $$y=k$$ on the ellipse equals $$2\sqrt{3}$$. 5. **Find the chord length at $$y=k$$:** Substitute $$y=k$$ into ellipse equation: $$\frac{x^2}{4} + \frac{k^2}{16} = 1 \implies \frac{x^2}{4} = 1 - \frac{k^2}{16} \implies x^2 = 4\left(1 - \frac{k^2}{16}\right) = 4 - \frac{k^2}{4}$$ 6. **Chord length $$L$$ is twice the positive $$x$$-value:** $$L = 2\sqrt{4 - \frac{k^2}{4}}$$ 7. **Set chord length equal to semi-focal distance:** $$2\sqrt{4 - \frac{k^2}{4}} = 2\sqrt{3}$$ 8. **Divide both sides by 2:** $$\cancel{2}\sqrt{4 - \frac{k^2}{4}} = \cancel{2}\sqrt{3} \implies \sqrt{4 - \frac{k^2}{4}} = \sqrt{3}$$ 9. **Square both sides:** $$4 - \frac{k^2}{4} = 3$$ 10. **Solve for $$k^2$$:** $$4 - 3 = \frac{k^2}{4} \implies 1 = \frac{k^2}{4} \implies k^2 = 4$$ 11. **Find $$k$$:** $$k = \pm 2$$ **Final answer:** The lines are $$y = 2$$ and $$y = -2$$.