1. **Problem statement:** Transform the conic given in polar coordinates $$r = \frac{4}{4 - 2 \cos \theta}$$ into rectangular coordinates and show it reduces to the ellipse equation $$\frac{(x - \frac{2}{3})^2}{(\frac{4}{3})^2} + \frac{y^2}{(\frac{2}{\sqrt{3}})^2} = 1$$. Then identify the conic and find its center, vertices, foci, major axis, eccentricity, and directrix. 2. **Recall the polar to rectangular coordinate relations:** $$x = r \cos \theta, \quad y = r \sin \theta$$ 3. **Start with the given polar equation:** $$r = \frac{4}{4 - 2 \cos \theta}$$ Multiply both sides by the denominator: $$r(4 - 2 \cos \theta) = 4$$ Distribute $r$: $$4r - 2r \cos \theta = 4$$ 4. **Express $r \cos \theta$ in terms of $x$:** Since $x = r \cos \theta$, substitute: $$4r - 2x = 4$$ 5. **Isolate $r$:** $$4r = 4 + 2x$$ $$r = \frac{4 + 2x}{4} = 1 + \frac{x}{2}$$ 6. **Recall $r = \sqrt{x^2 + y^2}$, so substitute:** $$\sqrt{x^2 + y^2} = 1 + \frac{x}{2}$$ 7. **Square both sides to eliminate the square root:** $$x^2 + y^2 = \left(1 + \frac{x}{2}\right)^2 = 1 + x + \frac{x^2}{4}$$ 8. **Bring all terms to one side:** $$x^2 + y^2 - 1 - x - \frac{x^2}{4} = 0$$ 9. **Combine like terms:** $$x^2 - \frac{x^2}{4} = \frac{3x^2}{4}$$ So, $$\frac{3x^2}{4} + y^2 - x - 1 = 0$$ 10. **Multiply entire equation by 4 to clear denominators:** $$3x^2 + 4y^2 - 4x - 4 = 0$$ 11. **Group $x$ and $y$ terms:** $$3x^2 - 4x + 4y^2 = 4$$ 12. **Complete the square for $x$ terms:** Factor out 3 from $x$ terms: $$3(x^2 - \frac{4}{3}x) + 4y^2 = 4$$ Complete the square inside parentheses: $$x^2 - \frac{4}{3}x = \left(x - \frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 = \left(x - \frac{2}{3}\right)^2 - \frac{4}{9}$$ 13. **Substitute back:** $$3\left(\left(x - \frac{2}{3}\right)^2 - \frac{4}{9}\right) + 4y^2 = 4$$ Distribute 3: $$3\left(x - \frac{2}{3}\right)^2 - \frac{12}{9} + 4y^2 = 4$$ Simplify $\frac{12}{9} = \frac{4}{3}$: $$3\left(x - \frac{2}{3}\right)^2 + 4y^2 = 4 + \frac{4}{3} = \frac{12}{3} + \frac{4}{3} = \frac{16}{3}$$ 14. **Divide entire equation by $\frac{16}{3}$ to get standard ellipse form:** $$\frac{3\left(x - \frac{2}{3}\right)^2}{\frac{16}{3}} + \frac{4y^2}{\frac{16}{3}} = 1$$ Simplify denominators: $$\frac{3}{\frac{16}{3}} = \frac{3 \times 3}{16} = \frac{9}{16}$$ $$\frac{4}{\frac{16}{3}} = \frac{4 \times 3}{16} = \frac{12}{16} = \frac{3}{4}$$ So: $$\frac{9}{16}\left(x - \frac{2}{3}\right)^2 + \frac{3}{4} y^2 = 1$$ 15. **Rewrite as:** $$\frac{(x - \frac{2}{3})^2}{\frac{16}{9}} + \frac{y^2}{\frac{4}{3}} = 1$$ 16. **Recognize the ellipse equation:** $$\frac{(x - \frac{2}{3})^2}{(\frac{4}{3})^2} + \frac{y^2}{(\frac{2}{\sqrt{3}})^2} = 1$$ 17. **Identify the conic:** This is an ellipse. 18. **Center coordinates:** $$(x, y) = \left(\frac{2}{3}, 0\right)$$ 19. **Vertices:** Along the major axis (x-axis), vertices are at: $$\left(\frac{2}{3} \pm \frac{4}{3}, 0\right) = \left(-\frac{2}{3}, 0\right) \text{ and } \left(2, 0\right)$$ 20. **Calculate eccentricity $e$:** $$a = \frac{4}{3}, \quad b = \frac{2}{\sqrt{3}}$$ $$c = \sqrt{a^2 - b^2} = \sqrt{\left(\frac{4}{3}\right)^2 - \left(\frac{2}{\sqrt{3}}\right)^2} = \sqrt{\frac{16}{9} - \frac{4}{3}} = \sqrt{\frac{16}{9} - \frac{12}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$$ $$e = \frac{c}{a} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{1}{2}$$ 21. **Foci coordinates:** $$\left(\frac{2}{3} \pm \frac{2}{3}, 0\right) = (0, 0) \text{ and } \left(\frac{4}{3}, 0\right)$$ 22. **Equation of major axis:** The major axis is along the horizontal line through the center: $$y = 0$$ 23. **Equation of directrices:** Directrices are vertical lines at distance $\frac{a}{e}$ from the center: $$x = \frac{2}{3} \pm \frac{a}{e} = \frac{2}{3} \pm \frac{\frac{4}{3}}{\frac{1}{2}} = \frac{2}{3} \pm \frac{4}{3} \times 2 = \frac{2}{3} \pm \frac{8}{3}$$ So, $$x = -2 \text{ and } x = \frac{10}{3}$$ 24. **Summary:** - The conic is an ellipse. - Center: $\left(\frac{2}{3}, 0\right)$ - Vertices: $\left(-\frac{2}{3}, 0\right)$ and $(2, 0)$ - Foci: $(0, 0)$ and $\left(\frac{4}{3}, 0\right)$ - Major axis: $y = 0$ - Eccentricity: $\frac{1}{2}$ - Directrices: $x = -2$ and $x = \frac{10}{3}$ This completes the transformation and identification of the conic.