1. **Problem statement:** Show that the equation $3x^2 - 4y^2 + 3x + 16y - 18 = 0$ represents a hyperbola. Then find its eccentricity and the equations of its asymptotes. 2. **Rewrite the equation:** Group $x$ and $y$ terms: $$3x^2 + 3x - 4y^2 + 16y = 18$$ 3. **Complete the square:** For $x$ terms: $$3x^2 + 3x = 3(x^2 + x) = 3\left(x^2 + x + \frac{1}{4} - \frac{1}{4}\right) = 3\left(\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}\right) = 3\left(x + \frac{1}{2}\right)^2 - \frac{3}{4}$$ For $y$ terms: $$-4y^2 + 16y = -4(y^2 - 4y) = -4\left(y^2 - 4y + 4 - 4\right) = -4\left((y - 2)^2 - 4\right) = -4(y - 2)^2 + 16$$ 4. **Substitute back:** $$3\left(x + \frac{1}{2}\right)^2 - \frac{3}{4} - 4(y - 2)^2 + 16 = 18$$ 5. **Simplify constants:** $$3\left(x + \frac{1}{2}\right)^2 - 4(y - 2)^2 + \left(-\frac{3}{4} + 16\right) = 18$$ $$3\left(x + \frac{1}{2}\right)^2 - 4(y - 2)^2 + \frac{61}{4} = 18$$ 6. **Isolate terms:** $$3\left(x + \frac{1}{2}\right)^2 - 4(y - 2)^2 = 18 - \frac{61}{4} = \frac{72}{4} - \frac{61}{4} = \frac{11}{4}$$ 7. **Divide both sides by $\frac{11}{4}$:** $$\frac{3\left(x + \frac{1}{2}\right)^2}{\frac{11}{4}} - \frac{4(y - 2)^2}{\frac{11}{4}} = 1$$ 8. **Simplify fractions:** $$\frac{3 \cdot 4}{11} \left(x + \frac{1}{2}\right)^2 - \frac{4 \cdot 4}{11} (y - 2)^2 = 1$$ $$\frac{12}{11} \left(x + \frac{1}{2}\right)^2 - \frac{16}{11} (y - 2)^2 = 1$$ 9. **Rewrite as:** $$\frac{\left(x + \frac{1}{2}\right)^2}{\frac{11}{12}} - \frac{(y - 2)^2}{\frac{11}{16}} = 1$$ This is the standard form of a hyperbola centered at $\left(-\frac{1}{2}, 2\right)$ with $a^2 = \frac{11}{12}$ and $b^2 = \frac{11}{16}$. 10. **Eccentricity $e$ of hyperbola:** $$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\frac{11}{16}}{\frac{11}{12}}} = \sqrt{1 + \frac{11}{16} \cdot \frac{12}{11}} = \sqrt{1 + \frac{12}{16}} = \sqrt{1 + \frac{3}{4}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}$$ 11. **Equations of asymptotes:** For hyperbola $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$, asymptotes are: $$y - k = \pm \frac{b}{a} (x - h)$$ Here, $$h = -\frac{1}{2}, \quad k = 2, \quad a = \sqrt{\frac{11}{12}}, \quad b = \sqrt{\frac{11}{16}}$$ Calculate $\frac{b}{a}$: $$\frac{b}{a} = \frac{\sqrt{\frac{11}{16}}}{\sqrt{\frac{11}{12}}} = \sqrt{\frac{11}{16} \cdot \frac{12}{11}} = \sqrt{\frac{12}{16}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$ So asymptotes: $$y - 2 = \pm \frac{\sqrt{3}}{2} \left(x + \frac{1}{2}\right)$$ --- **Summary for part (a):** - The equation represents a hyperbola. - Eccentricity $e = \frac{\sqrt{7}}{2}$. - Asymptotes: $$y - 2 = \pm \frac{\sqrt{3}}{2} \left(x + \frac{1}{2}\right)$$ --- 1. **Problem statement (part b):** Show that $r = \frac{7}{2(1 - \cos \theta)}$ is the polar form of a parabola. 2. **Recall polar conic form:** $$r = \frac{ed}{1 + e \cos \theta}$$ or $$r = \frac{ed}{1 - e \cos \theta}$$ where $e$ is eccentricity and $d$ is the distance from focus to directrix. 3. **Compare given equation:** $$r = \frac{7}{2(1 - \cos \theta)} = \frac{7/2}{1 - \cos \theta}$$ 4. **Identify parameters:** $$e = 1$$ (since denominator is $1 - \cos \theta$ and coefficient of $\cos \theta$ is 1) $$ed = \frac{7}{2}$$ 5. **Since $e=1$, the conic is a parabola.** **Summary for part (b):** The given polar equation represents a parabola with eccentricity $e=1$ and $ed=\frac{7}{2}$.